Reguli pentru integrarea generală a funcțiilor

REGULI PENTRU INTEGRAREA GENERALĂ A FUNCŢIILOR

$Pentru\ a\ real\ nenul:\int af(x)\,dx = a\int f(x)\,dx$

$\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx$

$\int f(x)g(x)\,dx = f(x)\int g(x)\,dx - \int \left(\int g(x)\,dx\right)\,d(f(x))$

FUNCŢII RAŢIONALE

$\int 1\,dx = x + C$

$\int x\,dx = \frac{x^2}{2}\ + C$

$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\qquad\mbox{ daca }n \ne -1$

$\int \frac{1}{x}\,dx = \ln{\left|x\right|} + C$

$\int {dx \over {x^2+a^2}} = {1 \over a} arctg {x \over a} + C$

$\int (ax + b)^n dx$	$= \frac{(ax + b)^{n+1}}{a(n + 1)} \qquad\mbox{(pentru } n\neq -1\mbox{)}\,\!$
$\int\frac{dx}{ax + b}$	$= \frac{1}{a}\ln\left\|ax + b\right\|$
$\int x(ax + b)^n dx$	$= \frac{a(n + 1)x - b}{a^2(n + 1)(n + 2)} (ax + b)^{n+1} \qquad\mbox{(pentru }n \not\in \{1, 2\}\mbox{)}$

$\int\frac{x dx}{ax + b}$	$= \frac{x}{a} - \frac{b}{a^2}\ln\left\|ax + b\right\|$
$\int\frac{x dx}{(ax + b)^2}$	$= \frac{b}{a^2(ax + b)} + \frac{1}{a^2}\ln\left\|ax + b\right\|$
$\int\frac{x dx}{(ax + b)^n}$	$= \frac{a(1 - n)x - b}{a^2(n - 1)(n - 2)(ax + b)^{n-1}} \qquad\mbox{(pentru } n\not\in \{1, 2\}\mbox{)}$

$\int\frac{x^2 dx}{ax + b}$	$= \frac{1}{a^3}\left(\frac{(ax + b)^2}{2} - 2b(ax + b) + b^2\ln\left\|ax + b\right\|\right)$
$\int\frac{x^2 dx}{(ax + b)^2}$	$= \frac{1}{a^3}\left(ax + b - 2b\ln\left\|ax + b\right\| - \frac{b^2}{ax + b}\right)$
$\int\frac{x^2 dx}{(ax + b)^3}$	$= \frac{1}{a^3}\left(\ln\left\|ax + b\right\| + \frac{2b}{ax + b} - \frac{b^2}{2(ax + b)^2}\right)$
$\int\frac{x^2 dx}{(ax + b)^n}$	$= \frac{1}{a^3}\left(-\frac{(ax + b)^{3-n}}{(n-3)} + \frac{2b (a + b)^{2-n}}{(n-2)} - \frac{b^2 (ax + b)^{1-n}}{(n - 1)}\right) \qquad\mbox{(pentru } n\not\in \{1, 2, 3\}\mbox{)}$

$\int\frac{dx}{x(ax + b)}$	$= -\frac{1}{b}\ln\left\|\frac{ax+b}{x}\right\|$
$\int\frac{dx}{x^2(ax+b)}$	$= -\frac{1}{bx} + \frac{a}{b^2}\ln\left\|\frac{ax+b}{x}\right\|$
$\int\frac{dx}{x^2(ax+b)^2}$	$= -a\left(\frac{1}{b^2(ax+b)} + \frac{1}{ab^2x} - \frac{2}{b^3}\ln\left\|\frac{ax+b}{x}\right\|\right)$
$\int\frac{dx}{x^2+a^2}$	$= \frac{1}{a}\arctan\frac{x}{a}\,\!$
$\int\frac{dx}{x^2-a^2} =$	$-\frac{1}{a}\,\mathrm{arctanh}\frac{x}{a} = \frac{1}{2a}\ln\frac{a-x}{a+x} \qquad\mbox{(pentru }\|x\| < \|a\|\mbox{)}\,\!$
	$-\frac{1}{a}\,\mathrm{arccoth}\frac{x}{a} = \frac{1}{2a}\ln\frac{x-a}{x+a} \qquad\mbox{(pentru }\|x\| > \|a\|\mbox{)}\,\!$

$\int\frac{dx}{ax^2+bx+c} =$	$\frac{2}{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}} \qquad\mbox{(pentru }4ac-b^2>0\mbox{)}$
	$\frac{2}{\sqrt{b^2-4ac}}\,\mathrm{artanh}\frac{2ax+b}{\sqrt{b^2-4ac}} = \frac{1}{\sqrt{b^2-4ac}}\ln\left\|\frac{2ax+b-\sqrt{b^2-4ac}}{2ax+b+\sqrt{b^2-4ac}}\right\| \qquad\mbox{(pentru }4ac-b^2<0\mbox{)}$
	$-\frac{2}{2ax+b}\qquad\mbox{(pentru }4ac-b^2=0\mbox{)}$
$\int\frac{x dx}{ax^2+bx+c}$	$= \frac{1}{2a}\ln\left\|ax^2+bx+c\right\|-\frac{b}{2a}\int\frac{dx}{ax^2+bx+c}$

$\int\frac{(mx+n) dx}{ax^2+bx+c} =$	$\frac{m}{2a}\ln\left\|ax^2+bx+c\right\|+\frac{2an-bm}{a\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}} \qquad\mbox{(pentru }4ac-b^2>0\mbox{)}$
	$\frac{m}{2a}\ln\left\|ax^2+bx+c\right\|+\frac{2an-bm}{a\sqrt{b^2-4ac}}\,\mathrm{artanh}\frac{2ax+b}{\sqrt{b^2-4ac}} \qquad\mbox{(pentru }4ac-b^2<0\mbox{)}$
	$\frac{m}{2a}\ln\left\|ax^2+bx+c\right\|-\frac{2an-bm}{a(2ax+b)} \,\,\,\,\,\,\,\,\,\, \qquad\mbox{(pentru }4ac-b^2=0\mbox{)}$

$\int\frac{dx}{(ax^2+bx+c)^n} = \frac{2ax+b}{(n-1)(4ac-b^2)(ax^2+bx+c)^{n-1}}+\frac{(2n-3)2a}{(n-1)(4ac-b^2)}\int\frac{dx}{(ax^2+bx+c)^{n-1}}\,\!$

$\int\frac{x dx}{(ax^2+bx+c)^n} = \frac{bx+2c}{(n-1)(4ac-b^2)(ax^2+bx+c)^{n-1}}-\frac{b(2n-3)}{(n-1)(4ac-b^2)}\int\frac{dx}{(ax^2+bx+c)^{n-1}}\,\!$

$\int\frac{dx}{x(ax^2+bx+c)} = \frac{1}{2c}\ln\left|\frac{x^2}{ax^2+bx+c}\right|-\frac{b}{2c}\int\frac{dx}{ax^2+bx+c}$

Orice funcție rațională poate fi integrată folosind ecuațiile de mai sus şi descompunerea parţială a funcţiei rationale

în sumă de funcții de forma:

$\frac{ex + f}{\left(ax^2+bx+c\right)^n}$ .

FUNCŢII IRAŢIONALE

$\int {dx \over \sqrt{a^2-x^2}} = \arcsin {x \over a} + C$

$\int {-dx \over \sqrt{a^2-x^2}} = \arccos {x \over a} + C$

$\int {dx \over x\sqrt{x^2-a^2}} = {1 \over a}\mbox{arcsec}\,{|x| \over a} + C$

INTEGRALE cu $r = \sqrt{x^2+a^2}$

$\int r \;dx = \frac{1}{2}\left(x r +a^2\,\ln\left(x+r\right)\right)$

$\int r^3 \;dx = \frac{1}{4}xr^3+\frac{1}{8}3a^2xr+\frac{3}{8}a^4\ln\left(x+r\right)$

$\int r^5 \; dx = \frac{1}{6}xr^5+\frac{5}{24}a^2xr^3+\frac{5}{16}a^4xr+\frac{5}{16}a^6\ln\left(x+r\right)$

$\int x r\;dx=\frac{r^3}{3}$

$\int x r^3\;dx=\frac{r^5}{5}$

$\int x r^{2n+1}\;dx=\frac{r^{2n+3}}{2n+3}$

$\int x^2 r\;dx= \frac{xr^3}{4}-\frac{a^2xr}{8}-\frac{a^4}{8}\ln\left(x+r\right)$

$\int x^2 r^3\;dx= \frac{xr^5}{6}-\frac{a^2xr^3}{24}-\frac{a^4xr}{16}-\frac{a^6}{16}\ln\left(x+r\right)$

$\int x^3 r \; dx = \frac{r^5}{5} - \frac{a^2 r^3}{3}$

$\int x^3 r^3 \; dx = \frac{r^7}{7}-\frac{a^2r^5}{5}$

$\int x^3 r^{2n+1} \; dx = \frac{r^{2n+5}}{2n+5} - \frac{a^3 r^{2n+3}}{2n+3}$

$\int x^4 r\;dx= \frac{x^3r^3}{6}-\frac{a^2xr^3}{8}+\frac{a^4xr}{16}+\frac{a^6}{16}\ln\left(x+r\right)$

$\int x^4 r^3\;dx= \frac{x^3r^5}{8}-\frac{a^2xr^5}{16}+\frac{a^4xr^3}{64}+\frac{3a^6xr}{128}+\frac{3a^8}{128}\ln\left(x+r\right)$

$\int x^5 r \; dx = \frac{r^7}{7} - \frac{2 a^2 r^5}{5} + \frac{a^4 r^3}{3}$

$\int x^5 r^3 \; dx = \frac{r^9}{9} - \frac{2 a^2 r^7}{7} + \frac{a^4 r^5}{5}$

$\int x^5 r^{2n+1} \; dx = \frac{r^{2n+7}}{2n+7} - \frac{2a^2r^{2n+5}}{2n+5}+\frac{a^4 r^{2n+3}}{2n+3}$

$\int\frac{r\;dx}{x} = r-a\ln\left|\frac{a+r}{x}\right| = r - a \sinh^{-1}\frac{a}{x}$

$\int\frac{r^3\;dx}{x} = \frac{r^3}{3}+a^2r-a^3\ln\left|\frac{a+r}{x}\right|$

$\int\frac{r^5\;dx}{x} = \frac{r^5}{5}+\frac{a^2r^3}{3}+a^4r-a^5\ln\left|\frac{a+r}{x}\right|$

$\int\frac{r^7\;dx}{x} = \frac{r^7}{7}+\frac{a^2r^5}{5}+\frac{a^4r^3}{3}+a^6r-a^7\ln\left|\frac{a+r}{x}\right|$

$\int\frac{dx}{r} = \sinh^{-1}\frac{x}{a} = \ln\left|x+r\right|$

$\int\frac{dx}{r^3} = \frac{x}{a^2r}$

$\int\frac{x\,dx}{r} = r$

$\int\frac{x\,dx}{r^3} = -\frac{1}{r}$

$\int\frac{x^2\;dx}{r} = \frac{x}{2}r-\frac{a^2}{2}\,\sinh^{-1}\frac{x}{a} = \frac{x}{2}r-\frac{a^2}{2}\ln\left|x+r\right|$

$\int\frac{dx}{xr} = -\frac{1}{a}\,\sinh^{-1}\frac{a}{x} = -\frac{1}{a}\ln\left|\frac{a+r}{x}\right|$

INTEGRALE cu $s = \sqrt{x^2-a^2}$

Se presupune $($ $x$ $2$ $>$ $a$ $2$ $)$ , pentru $($ $x$ $2$ $<$ $a$ $2$ $)$ , vezi următoarea secțiune:

$\int xs\;dx = \frac{1}{3}s^3$

$\int\frac{s\;dx}{x} = s - a\cos^{-1}\left|\frac{a}{x}\right|$

$\int\frac{dx}{s} = \int\frac{dx}{\sqrt{x^2-a^2}} =\ln\left|\frac{x+s}{a}\right|$

$\ln\left|\frac{x+s}{a}\right| =\mathrm{sgn}(x)\cosh^{-1}\left|\frac{x}{a}\right| =\frac{1}{2}\ln\left(\frac{x+s}{x-s}\right)$ , unde se consideră valoarea pozitivă a lui $\cosh^{-1}\left|\frac{x}{a}\right|$

$\int\frac{x\;dx}{s} = s$

$\int\frac{x\;dx}{s^3} = -\frac{1}{s}$

$\int\frac{x\;dx}{s^5} = -\frac{1}{3s^3}$

$\int\frac{x\;dx}{s^7} = -\frac{1}{5s^5}$

$\int\frac{x\;dx}{s^{2n+1}} = -\frac{1}{(2n-1)s^{2n-1}}$

$\int\frac{x^{2m}\;dx}{s^{2n+1}} = -\frac{1}{2n-1}\frac{x^{2m-1}}{s^{2n-1}}+\frac{2m-1}{2n-1}\int\frac{x^{2m-2}\;dx}{s^{2n-1}}$

$\int\frac{x^2\;dx}{s} = \frac{xs}{2}+\frac{a^2}{2}\ln\left|\frac{x+s}{a}\right|$

$\int\frac{x^2\;dx}{s^3} = -\frac{x}{s}+\ln\left|\frac{x+s}{a}\right|$

$\int\frac{x^4\;dx}{s} = \frac{x^3s}{4}+\frac{3}{8}a^2xs+\frac{3}{8}a^4\ln\left|\frac{x+s}{a}\right|$

$\int\frac{x^4\;dx}{s^3} = \frac{xs}{2}-\frac{a^2x}{s}+\frac{3}{2}a^2\ln\left|\frac{x+s}{a}\right|$

$\int\frac{x^4\;dx}{s^5} = -\frac{x}{s}-\frac{1}{3}\frac{x^3}{s^3}+\ln\left|\frac{x+s}{a}\right|$

$\int\frac{x^{2m}\;dx}{s^{2n+1}} = (-1)^{n-m}\frac{1}{a^{2(n-m)}}\sum_{i=0}^{n-m-1}\frac{1}{2(m+i)+1}{n-m-1 \choose i}\frac{x^{2(m+i)+1}}{s^{2(m+i)+1}}\qquad\mbox{(}n>m\ge0\mbox{)}$

$\int\frac{dx}{s^3}=-\frac{1}{a^2}\frac{x}{s}$

$\int\frac{dx}{s^5}=\frac{1}{a^4}\left[\frac{x}{s}-\frac{1}{3}\frac{x^3}{s^3}\right]$

$\int\frac{dx}{s^7} =-\frac{1}{a^6}\left[\frac{x}{s}-\frac{2}{3}\frac{x^3}{s^3}+\frac{1}{5}\frac{x^5}{s^5}\right]$

$\int\frac{dx}{s^9} =\frac{1}{a^8}\left[\frac{x}{s}-\frac{3}{3}\frac{x^3}{s^3}+\frac{3}{5}\frac{x^5}{s^5}-\frac{1}{7}\frac{x^7}{s^7}\right]$

$\int\frac{x^2\;dx}{s^5}=-\frac{1}{a^2}\frac{x^3}{3s^3}$

$\int\frac{x^2\;dx}{s^7} = \frac{1}{a^4}\left[\frac{1}{3}\frac{x^3}{s^3}-\frac{1}{5}\frac{x^5}{s^5}\right]$

$\int\frac{x^2\;dx}{s^9} = -\frac{1}{a^6}\left[\frac{1}{3}\frac{x^3}{s^3}-\frac{2}{5}\frac{x^5}{s^5}+\frac{1}{7}\frac{x^7}{s^7}\right]$

INTEGRALE cu $t = \sqrt{a^2-x^2}$

$\int t \;dx = \frac{1}{2}\left(xt+a^2\sin^{-1}\frac{x}{a}\right) \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int xt\;dx = -\frac{1}{3} t^3 \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int\frac{t\;dx}{x} = t-a\ln\left|\frac{a+t}{x}\right| \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int\frac{dx}{t} = \sin^{-1}\frac{x}{a} \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int\frac{x^2\;dx}{t} = -\frac{x}{2}t+\frac{a^2}{2}\sin^{-1}\frac{x}{a} \qquad\mbox{(}|x|\leq|a|\mbox{)}$

$\int t\;dx = \frac{1}{2}\left(xt-\sgn x\,\cosh^{-1}\left|\frac{x}{a}\right|\right) \qquad\mbox{(for }|x|\ge|a|\mbox{)}$

INTEGRALE cu $R^{1/2} = \sqrt{ax^2+bx+c}$

$\int\frac{dx}{\sqrt{ax^2+bx+c}} = \frac{1}{\sqrt{a}}\ln\left|2\sqrt{a R}+2ax+b\right| \qquad \mbox{(for }a>0\mbox{)}$

$\int\frac{dx}{\sqrt{ax^2+bx+c}} = \frac{1}{\sqrt{a}}\,\sinh^{-1}\frac{2ax+b}{\sqrt{4ac-b^2}} \qquad \mbox{(for }a>0\mbox{, }4ac-b^2>0\mbox{)}$

$\int\frac{dx}{\sqrt{ax^2+bx+c}} = \frac{1}{\sqrt{a}}\ln|2ax+b| \quad \mbox{(for }a>0\mbox{, }4ac-b^2=0\mbox{)}$

$\int\frac{dx}{\sqrt{ax^2+bx+c}} = -\frac{1}{\sqrt{-a}}\arcsin\frac{2ax+b}{\sqrt{b^2-4ac}} \qquad \mbox{(for }a<0\mbox{, }4ac-b^2<0\mbox{)}$

$\int\frac{dx}{\sqrt{(ax^2+bx+c)^{3}}} = \frac{4ax+2b}{(4ac-b^2)\sqrt{R}}$

$\int\frac{dx}{\sqrt{(ax^2+bx+c)^{5}}} = \frac{4ax+2b}{3(4ac-b^2)\sqrt{R}}\left(\frac{1}{R}+\frac{8a}{4ac-b^2}\right)$

$\int\frac{dx}{\sqrt{(ax^2+bx+c)^{2n+1}}} = \frac{4ax+2b}{(2n-1)(4ac-b^2)R^{(2n-1)/2}}+\frac{8a(n-1)}{(2n-1)(4ac-b^2)} \int\frac{dx}{R^{(2n-1)/2}}$

$\int\frac{x\;dx}{\sqrt{ax^2+bx+c}} = \frac{\sqrt{R}}{a}-\frac{b}{2a}\int\frac{dx}{\sqrt{R}}$

$\int\frac{x\;dx}{\sqrt{(ax^2+bx+c)^3}} = -\frac{2bx+4c}{(4ac-b^2)\sqrt{R}}$

$\int\frac{x\;dx}{\sqrt{(ax^2+bx+c)^{2n+1}}} = -\frac{1}{(2n-1)aR^{(2n-1)/2}}-\frac{b}{2a}\int\frac{dx}{R^{(2n+1)/2}}$

$\int\frac{dx}{x\sqrt{ax^2+bx+c}}=-\frac{1}{\sqrt{c}}\ln\left(\frac{2\sqrt{c R}+bx+2c}{x}\right)$

$\int\frac{dx}{x\sqrt{ax^2+bx+c}}=-\frac{1}{\sqrt{c}}\sinh^{-1}\left(\frac{bx+2c}{|x|\sqrt{4ac-b^2}}\right)$

INTEGRALE cu $R^{1/2} = \sqrt{ax+b}$

$\int \frac{dx}{x\sqrt{ax + b}}\,=\,\frac{-2}{\sqrt{b}}\tanh^{-1}{\sqrt{\frac{ax + b}{b}}}$

$\int\frac{\sqrt{ax + b}}{x}\,dx\;=\;2\left(\sqrt{ax + b} - \sqrt{b}\tanh^{-1}{\sqrt{\frac{ax + b}{b}}}\right)$

$\int\frac{x^n}{\sqrt{ax + b}}\,dx\;=\;\frac{2}{a(2n+1)} \left(x^{n}\sqrt{ax + b} - bn\int\frac{x^{n-1}}{\sqrt{ax + b}}\right)$

$\int x^n \sqrt{ax + b}\,dx \; = \; \frac{2}{2n +1}\left(x^{n+1} \sqrt{ax + b} + bx^{n} \sqrt{ax + b} - nb\int x^{n-1}\sqrt{ax + b}\,dx \right)$

FUNCŢII LOGARITMICE

$\int \ln {x}\,dx = x (\ln {x}-1) + C$

$\int \log_b {x}\,dx = x\log_b {x} + x\log_b {e} + C$

$\int\ln cx\;dx = x\ln cx - x$

$\int\ln (ax + b)\;dx = x\ln (ax +b) - x + \frac{b}{a}\ln (ax + b)$

$\int (\ln x)^2\; dx = x(\ln x)^2 - 2x\ln x + 2x$

$\int (\ln cx)^n\; dx = x(\ln cx)^n - n\int (\ln cx)^{n-1} dx$

$\int \frac{dx}{\ln x} = \ln|\ln x| + \ln x + \sum^\infty_{i=2}\frac{(\ln x)^i}{i\cdot i!}$

$\int \frac{dx}{(\ln x)^n} = -\frac{x}{(n-1)(\ln x)^{n-1}} + \frac{1}{n-1}\int\frac{dx}{(\ln x)^{n-1}} \qquad\mbox{(pentru }n\neq 1\mbox{)}$

$\int x^m\ln x\;dx = x^{m+1}\left(\frac{\ln x}{m+1}-\frac{1}{(m+1)^2}\right) \qquad\mbox{(pentru }m\neq -1\mbox{)}$

$\int x^m (\ln x)^n\; dx = \frac{x^{m+1}(\ln x)^n}{m+1} - \frac{n}{m+1}\int x^m (\ln x)^{n-1} dx \qquad\mbox{(pentru }m\neq -1\mbox{)}$

$\int \frac{(\ln x)^n\; dx}{x} = \frac{(\ln x)^{n+1}}{n+1} \qquad\mbox{(pentru }n\neq -1\mbox{)}$

$\int \frac{\ln x\,dx}{x^m} = -\frac{\ln x}{(m-1)x^{m-1}}-\frac{1}{(m-1)^2 x^{m-1}} \qquad\mbox{(pentru }m\neq 1\mbox{)}$

$\int \frac{(\ln x)^n\; dx}{x^m} = -\frac{(\ln x)^n}{(m-1)x^{m-1}} + \frac{n}{m-1}\int\frac{(\ln x)^{n-1} dx}{x^m} \qquad\mbox{(pentru }m\neq 1\mbox{)}$

$\int \frac{x^m\; dx}{(\ln x)^n} = -\frac{x^{m+1}}{(n-1)(\ln x)^{n-1}} + \frac{m+1}{n-1}\int\frac{x^m dx}{(\ln x)^{n-1}} \qquad\mbox{(pentru }n\neq 1\mbox{)}$

$\int \frac{dx}{x\ln x} = \ln|\ln x|$

$\int \frac{dx}{x^n\ln x} = \ln|\ln x| + \sum^\infty_{i=1} (-1)^i\frac{(n-1)^i(\ln x)^i}{i\cdot i!}$

$\int \frac{dx}{x (\ln x)^n} = -\frac{1}{(n-1)(\ln x)^{n-1}} \qquad\mbox{(pentru }n\neq 1\mbox{)}$

$\int \sin (\ln x)\;dx = \frac{x}{2}(\sin (\ln x) - \cos (\ln x))$

$\int \cos (\ln x)\;dx = \frac{x}{2}(\sin (\ln x) + \cos (\ln x))$

$\int e^x (x \ln x - x - \frac{1}{x})\;dx = e^x (x \ln x - x - \ln x)$

FUNCŢII EXPONENŢIALE

$\int e^x\,dx = e^x + C$

$\int a^x\,dx = \frac{a^x}{\ln{a}} + C$

$\int e^{cx}\;dx = \frac{1}{c} e^{cx}$

$\int a^{cx}\;dx = \frac{1}{c \ln a} a^{cx} \qquad\mbox{(pentru } a > 0,\mbox{ }a \ne 1\mbox{)}$

$\int xe^{cx}\; dx = \frac{e^{cx}}{c^2}(cx-1)$

$\int x^2 e^{cx}\;dx = e^{cx}\left(\frac{x^2}{c}-\frac{2x}{c^2}+\frac{2}{c^3}\right)$

$\int x^n e^{cx}\; dx = \frac{1}{c} x^n e^{cx} - \frac{n}{c}\int x^{n-1} e^{cx} dx$

$\int\frac{e^{cx}\; dx}{x} = \ln|x| +\sum_{i=1}^\infty\frac{(cx)^i}{i\cdot i!}$

$\int\frac{e^{cx}\; dx}{x^n} = \frac{1}{n-1}\left(-\frac{e^{cx}}{x^{n-1}}+c\int\frac{e^{cx} }{x^{n-1}}\,dx\right) \qquad\mbox{(pentru }n\neq 1\mbox{)}$

$\int e^{cx}\ln x\; dx = \frac{1}{c}e^{cx}\ln|x|-\operatorname{Ei}\,(cx)$

$\int e^{cx}\sin bx\; dx = \frac{e^{cx}}{c^2+b^2}(c\sin bx - b\cos bx)$

$\int e^{cx}\cos bx\; dx = \frac{e^{cx}}{c^2+b^2}(c\cos bx + b\sin bx)$

$\int e^{cx}\sin^n x\; dx = \frac{e^{cx}\sin^{n-1} x}{c^2+n^2}(c\sin x-n\cos x)+\frac{n(n-1)}{c^2+n^2}\int e^{cx}\sin^{n-2} x\;dx$

$\int e^{cx}\cos^n x\; dx = \frac{e^{cx}\cos^{n-1} x}{c^2+n^2}(c\cos x+n\sin x)+\frac{n(n-1)}{c^2+n^2}\int e^{cx}\cos^{n-2} x\;dx$

$\int x e^{c x^2 }\; dx= \frac{1}{2c} \; e^{c x^2}$

$\int {1 \over \sigma\sqrt{2\pi} }\,e^{-{(x-\mu )^2 / 2\sigma^2}}\; dx= \frac{1}{2 \sigma} (1 + \mbox{erf}\,\frac{x-\mu}{\sigma \sqrt{2}})$

$\int e^{x^2}\,dx = e^{x^2}\left( \sum_{j=0}^{n-1}c_{2j}\,\frac{1}{x^{2j+1}} \right )+(2n-1)c_{2n-2} \int \frac{e^{x^2}}{x^{2n}}\;dx \quad \mbox{pentru } n > 0,$

unde $c_{2j}=\frac{ 1 \cdot 3 \cdot 5 \cdots (2j-1)}{2^{j+1}}=\frac{2j\,!}{j!\, 2^{2j+1}} \ .$

$\int_{-\infty}^{\infty} e^{-ax^2}\,dx=\sqrt{\pi \over a}$ (Integrala gaussiană)

$\int_{-\infty}^{\infty} x e^{-a(x-b)^2}\,dx=b \sqrt{\pi \over a}$

$\int_{-\infty}^{\infty} x^2 e^{-ax^2}\,dx=\frac{1}{2} \sqrt{\pi \over a^3}$

$\int_{0}^{\infty} x^{2n} e^{-{x^2}/{a^2}}\,dx=\sqrt{\pi} {(2n)! \over {n!}} {\left (\frac{a}{2} \right)}^{2n + 1}$

$\int_{-\infty}^{\infty} e^{-{x^2}/{a^2}} \cos bx\,dx=a \sqrt{\pi}(\sin{a^2 b^2 \over 4}+\cos{a^2 b^2\over 4})$

$\int_{0}^{2 \pi} e^{x \cos \theta} d \theta = 2 \pi I_{0}(x)$ ( $I$ $0$ este funcția Bessel de speța I modificată)

$\int_{0}^{2 \pi} e^{x \cos \theta + y \sin \theta} d \theta = 2 \pi I_{0} \left( \sqrt{x^2 + y^2} \right)$

$\int_{0}^{\infty} x^a e^{-bx} dx = \frac{a!}{b^{a+1}}$

FUNCŢII TRIGONOMETRICE

$\int \sin{x}\, dx = -\cos{x} + C$

$\int \cos{x}\, dx = \sin{x} + C$

$\int \tan{x} \, dx = -\ln{\left| \cos {x} \right|} + C$

$\int \cot{x} \, dx = \ln{\left| \sin{x} \right|} + C$

$\int \sec{x} \, dx = \ln{\left| \sec{x} + \tan{x}\right|} + C$

$\int \csc{x} \, dx = -\ln{\left| \csc{x} + \cot{x}\right|} + C$

$\int \sec^2 x \, dx = \tan x + C$

$\int \csc^2 x \, dx = -\cot x + C$

$\int \sec{x} \, \tan{x} \, dx = \sec{x} + C$

$\int \csc{x} \, \cot{x} \, dx = - \csc{x} + C$

$\int \sin^2 x \, dx = \frac{1}{2}(x - \sin x \cos x) + C$

$\int \cos^2 x \, dx = \frac{1}{2}(x + \sin x \cos x) + C$

$\int \sin^n x \, dx = - \frac{\sin^{n-1} {x} \cos {x}}{n} + \frac{n-1}{n} \int \sin^{n-2}{x} \, dx$

$\int \cos^n x \, dx = \frac{\cos^{n-1} {x} \sin {x}}{n} + \frac{n-1}{n} \int \cos^{n-2}{x} \, dx$

$\int \arctan{x} \, dx = x \, \arctan{x} - \frac{1}{2} \ln{\left| 1 + x^2\right|} + C$

INTEGRALE de funcţii trigonometrice ce conţin numai sin

unde c este o constantă:

$\int\sin cx\;dx = -\frac{1}{c}\cos cx\,\!$

$\int\sin^n {cx}\;dx = -\frac{\sin^{n-1} cx\cos cx}{nc} + \frac{n-1}{n}\int\sin^{n-2} cx\;dx \qquad\mbox{(pentru }n>0\mbox{)}\,\!$

$\int\sin^2 {cx}\;dx = \frac{x}{2} - \frac{1}{4c} sin 2cx \!$

$\int\sqrt{1 - \sin{x}}\,dx = \int\sqrt{\operatorname{cvs}{x}}\,dx = 2 \frac{\cos{\frac{x}{2}} + \sin{\frac{x}{2}}}{\cos{\frac{x}{2}} - \sin{\frac{x}{2}}} \sqrt{\operatorname{cvs}{x}} = 2\sqrt{1 + \sin{x}}$

unde cvs{x} este funcția Coversinus

$\int x\sin cx\;dx = \frac{\sin cx}{c^2}-\frac{x\cos cx}{c}\,\!$

$\int x^n\sin cx\;dx = -\frac{x^n}{c}\cos cx+\frac{n}{c}\int x^{n-1}\cos cx\;dx \qquad\mbox{(pentru }n>0\mbox{)}\,\!$

$\int_{\frac{-a}{2}}^{\frac{a}{2}} x^2\sin^2 {\frac{n\pi x}{a}}\;dx = \frac{a^3(n^2\pi^2-6)}{24n^2\pi^2} \qquad\mbox{(pentru }n=2,4,6...\mbox{)}\,\!$

$\int\frac{\sin cx}{x} dx = \sum_{i=0}^\infty (-1)^i\frac{(cx)^{2i+1}}{(2i+1)\cdot (2i+1)!}\,\!$

$\int\frac{\sin cx}{x^n} dx = -\frac{\sin cx}{(n-1)x^{n-1}} + \frac{c}{n-1}\int\frac{\cos cx}{x^{n-1}} dx\,\!$

$\int\frac{dx}{\sin cx} = \frac{1}{c}\ln \left|\tan\frac{cx}{2}\right|$

$\int\frac{dx}{\sin^n cx} = \frac{\cos cx}{c(1-n) \sin^{n-1} cx}+\frac{n-2}{n-1}\int\frac{dx}{\sin^{n-2}cx} \qquad\mbox{(pentru }n>1\mbox{)}\,\!$

$\int\frac{dx}{1\pm\sin cx} = \frac{1}{c}\tan\left(\frac{cx}{2}\mp\frac{\pi}{4}\right)$

$\int\frac{x\;dx}{1+\sin cx} = \frac{x}{c}\tan\left(\frac{cx}{2} - \frac{\pi}{4}\right)+\frac{2}{c^2}\ln\left|\cos\left(\frac{cx}{2}-\frac{\pi}{4}\right)\right|$

$\int\frac{x\;dx}{1-\sin cx} = \frac{x}{c}\cot\left(\frac{\pi}{4} - \frac{cx}{2}\right)+\frac{2}{c^2}\ln\left|\sin\left(\frac{\pi}{4}-\frac{cx}{2}\right)\right|$

$\int\frac{\sin cx\;dx}{1\pm\sin cx} = \pm x+\frac{1}{c}\tan\left(\frac{\pi}{4}\mp\frac{cx}{2}\right)$

$\int\sin c_1x\sin c_2x\;dx = \frac{\sin(c_1-c_2)x}{2(c_1-c_2)}-\frac{\sin(c_1+c_2)x}{2(c_1+c_2)} \qquad\mbox{(pentru }|c_1|\neq|c_2|\mbox{)}\,\!$

INTEGRALE de funcţii trigonometrice ce conţin numai cos

$\int\cos cx\;dx = \frac{1}{c}\sin cx\,\!$

$\int\cos^n cx\;dx = \frac{\cos^{n-1} cx\sin cx}{nc} + \frac{n-1}{n}\int\cos^{n-2} cx\;dx \qquad\mbox{(pentru }n>0\mbox{)}\,\!$

$\int x\cos cx\;dx = \frac{\cos cx}{c^2} + \frac{x\sin cx}{c}\,\!$

$\int x^n\cos cx\;dx = \frac{x^n\sin cx}{c} - \frac{n}{c}\int x^{n-1}\sin cx\;dx\,\!$

$\int_{\frac{-a}{2}}^{\frac{a}{2}} x^2\cos^2 {\frac{n\pi x}{a}}\;dx = \frac{a^3(n^2\pi^2-6)}{24n^2\pi^2} \qquad\mbox{(pentru }n=1,3,5...\mbox{)}\,\!$

$\int\frac{\cos cx}{x} dx = \ln|cx|+\sum_{i=1}^\infty (-1)^i\frac{(cx)^{2i}}{2i\cdot(2i)!}\,\!$

$\int\frac{\cos cx}{x^n} dx = -\frac{\cos cx}{(n-1)x^{n-1}}-\frac{c}{n-1}\int\frac{\sin cx}{x^{n-1}} dx \qquad\mbox{(pentru }n\neq 1\mbox{)}\,\!$

$\int\frac{dx}{\cos cx} = \frac{1}{c}\ln\left|\tan\left(\frac{cx}{2}+\frac{\pi}{4}\right)\right|$

$\int\frac{dx}{\cos^n cx} = \frac{\sin cx}{c(n-1) cos^{n-1} cx} + \frac{n-2}{n-1}\int\frac{dx}{\cos^{n-2} cx} \qquad\mbox{(for }n>1\mbox{)}\,\!$

$\int\frac{dx}{1+\cos cx} = \frac{1}{c}\tan\frac{cx}{2}\,\!$

$\int\frac{dx}{1-\cos cx} = -\frac{1}{c}\cot\frac{cx}{2}\,\!$

$\int\frac{x\;dx}{1+\cos cx} = \frac{x}{c}\tan\frac{cx}{2} + \frac{2}{c^2}\ln\left|\cos\frac{cx}{2}\right|$

$\int\frac{x\;dx}{1-\cos cx} = -\frac{x}{c}\cot\frac{cx}{2}+\frac{2}{c^2}\ln\left|\sin\frac{cx}{2}\right|$

$\int\frac{\cos cx\;dx}{1+\cos cx} = x - \frac{1}{c}\tan\frac{cx}{2}\,\!$

$\int\frac{\cos cx\;dx}{1-\cos cx} = -x-\frac{1}{c}\cot\frac{cx}{2}\,\!$

$\int\cos c_1x\cos c_2x\;dx = \frac{\sin(c_1-c_2)x}{2(c_1-c_2)}+\frac{\sin(c_1+c_2)x}{2(c_1+c_2)} \qquad\mbox{(pentru }|c_1|\neq|c_2|\mbox{)}\,\!$

INTEGRALE de funcţii trigonometrice ce conţin numai tan

$\int\tan cx\;dx = -\frac{1}{c}\ln|\cos cx|\,\! = \frac{1}{c}\ln|\sec cx|\,\!$

$\int\frac{dx}{\tan cx} = \frac{1}{c}\ln|\sin cx|\,\!$

$\int\tan^n cx\;dx = \frac{1}{c(n-1)}\tan^{n-1} cx-\int\tan^{n-2} cx\;dx \qquad\mbox{(pentru }n\neq 1\mbox{)}\,\!$

$\int\frac{dx}{\tan cx + 1} = \frac{x}{2} + \frac{1}{2c}\ln|\sin cx + \cos cx|\,\!$

$\int\frac{dx}{\tan cx - 1} = -\frac{x}{2} + \frac{1}{2c}\ln|\sin cx - \cos cx|\,\!$

$\int\frac{\tan cx\;dx}{\tan cx + 1} = \frac{x}{2} - \frac{1}{2c}\ln|\sin cx + \cos cx|\,\!$

$\int\frac{\tan cx\;dx}{\tan cx - 1} = \frac{x}{2} + \frac{1}{2c}\ln|\sin cx - \cos cx|\,\!$

INTEGRALE de funcţii trigonometrice ce conţin numai sec

$\int \sec{cx} \, dx = \frac{1}{c}\ln{\left| \sec{cx} + \tan{cx}\right|}$

$\int \sec^n{cx} \, dx = \frac{\sec^{n-1}{cx} \sin {cx}}{c(n-1)} \,+\, \frac{n-2}{n-1}\int \sec^{n-2}{cx} \, dx \qquad \mbox{ (pentru }n \ne 1\mbox{)}\,\!$

$\int \frac{dx}{\sec{x} + 1} = x - \tan{\frac{x}{2}}$

INTEGRALE de funcţii trigonometrice ce conţin numai csc

$\int \csc{cx} \, dx = -\frac{1}{c}\ln{\left| \csc{cx} + \cot{cx}\right|}$

$\int \csc^n{cx} \, dx = -\frac{\csc^{n-1}{cx} \cos{cx}}{c(n-1)} \,+\, \frac{n-2}{n-1}\int \csc^{n-2}{cx} \, dx \qquad \mbox{ (pentru }n \ne 1\mbox{)}\,\!$

INTEGRALE de funcţii trigonometrice ce conţin numai cot

$\int\cot cx\;dx = \frac{1}{c}\ln|\sin cx|\,\!$

$\int\cot^n cx\;dx = -\frac{1}{c(n-1)}\cot^{n-1} cx - \int\cot^{n-2} cx\;dx \qquad\mbox{(pentru }n\neq 1\mbox{)}\,\!$

$\int\frac{dx}{1 + \cot cx} = \int\frac{\tan cx\;dx}{\tan cx+1}\,\!$

$\int\frac{dx}{1 - \cot cx} = \int\frac{\tan cx\;dx}{\tan cx-1}\,\!$

INTEGRALE funcţii trigonometrice ce conţn atât sin cât şi cos

$\int\frac{dx}{\cos cx\pm\sin cx} = \frac{1}{c\sqrt{2}}\ln\left|\tan\left(\frac{cx}{2}\pm\frac{\pi}{8}\right)\right|$

$\int\frac{dx}{(\cos cx\pm\sin cx)^2} = \frac{1}{2c}\tan\left(cx\mp\frac{\pi}{4}\right)$

$\int\frac{dx}{(\cos x + \sin x)^n} = \frac{1}{n-1}\left(\frac{\sin x - \cos x}{(\cos x + \sin x)^{n - 1}} - 2(n - 2)\int\frac{dx}{(\cos x + \sin x)^{n-2}} \right)$

$\int\frac{\cos cx\;dx}{\cos cx + \sin cx} = \frac{x}{2} + \frac{1}{2c}\ln\left|\sin cx + \cos cx\right|$

$\int\frac{\cos cx\;dx}{\cos cx - \sin cx} = frac{x}{2} - \frac{1}{2c}\ln\left|\sin cx - \cos cx\right|$

$\int\frac{\sin cx\;dx}{\cos cx + \sin cx} = \frac{x}{2} - \frac{1}{2c}\ln\left|\sin cx + \cos cx\right|$

$\int\frac{\sin cx\;dx}{\cos cx - \sin cx} = -\frac{x}{2} - \frac{1}{2c}\ln\left|\sin cx - \cos cx\right|$

$\int\frac{\cos cx\;dx}{\sin cx(1+\cos cx)} = -\frac{1}{4c}\tan^2\frac{cx}{2}+\frac{1}{2c}\ln\left|\tan\frac{cx}{2}\right|$

$\int\frac{\cos cx\;dx}{\sin cx(1+-\cos cx)} = -\frac{1}{4c}\cot^2\frac{cx}{2}-\frac{1}{2c}\ln\left|\tan\frac{cx}{2}\right|$

$\int\frac{\sin cx\;dx}{\cos cx(1+\sin cx)} = \frac{1}{4c}\cot^2\left(\frac{cx}{2}+\frac{\pi}{4}\right)+\frac{1}{2c}\ln\left|\tan\left(\frac{cx}{2}+\frac{\pi}{4}\right)\right|$

$\int\frac{\sin cx\;dx}{\cos cx(1-\sin cx)} = \frac{1}{4c}\tan^2\left(\frac{cx}{2}+\frac{\pi}{4}\right)-\frac{1}{2c}\ln\left|\tan\left(\frac{cx}{2}+\frac{\pi}{4}\right)\right|$

$\int\sin cx\cos cx\;dx = \frac{1}{2c}\sin^2 cx\,\!$

$\int\sin c_1x\cos c_2x\;dx = -\frac{\cos(c_1+c_2)x}{2(c_1+c_2)}-\frac{\cos(c_1-c_2)x}{2(c_1-c_2)} \qquad\mbox{(pentru }|c_1|\neq|c_2|\mbox{)}\,\!$

$\int\sin^n cx\cos cx\;dx = \frac{1}{c(n+1)}\sin^{n+1} cx \qquad\mbox{(pentru }n\neq 1\mbox{)}\,\!$

$\int\sin cx\cos^n cx\;dx = -\frac{1}{c(n+1)}\cos^{n+1} cx \qquad\mbox{(pentru }n\neq 1\mbox{)}\,\!$

$\int\sin^n cx\cos^m cx\;dx = -\frac{\sin^{n-1} cx\cos^{m+1} cx}{c(n+m)}+\frac{n-1}{n+m}\int\sin^{n-2} cx\cos^m cx\;dx \qquad\mbox{(pentru }m,n>0\mbox{)}\,\!$

also: $\int\sin^n cx\cos^m cx\;dx = \frac{\sin^{n+1} cx\cos^{m-1} cx}{c(n+m)} + \frac{m-1}{n+m}\int\sin^n cx\cos^{m-2} cx\;dx \qquad\mbox{(pentru }m,n>0\mbox{)}\,\!$

$\int\frac{dx}{\sin cx\cos cx} = \frac{1}{c}\ln\left|\tan cx\right|$

$\int\frac{dx}{\sin cx\cos^n cx} = \frac{1}{c(n-1)\cos^{n-1} cx}+\int\frac{dx}{\sin cx\cos^{n-2} cx} \qquad\mbox{(pentru }n\neq 1\mbox{)}\,\!$

$\int\frac{dx}{\sin^n cx\cos cx} = -\frac{1}{c(n-1)\sin^{n-1} cx}+\int\frac{dx}{\sin^{n-2} cx\cos cx} \qquad\mbox{(pentru }n\neq 1\mbox{)}\,\!$

$\int\frac{\sin cx\;dx}{\cos^n cx} = \frac{1}{c(n-1)\cos^{n-1} cx} \qquad\mbox{(pentru }n\neq 1\mbox{)}\,\!$

$\int\frac{\sin^2 cx\;dx}{\cos cx} = -\frac{1}{c}\sin cx+\frac{1}{c}\ln\left|\tan\left(\frac{\pi}{4}+\frac{cx}{2}\right)\right|$

$\int\frac{\sin^2 cx\;dx}{\cos^n cx} = \frac{\sin cx}{c(n-1)\cos^{n-1}cx}-\frac{1}{n-1}\int\frac{dx}{\cos^{n-2}cx} \qquad\mbox{(pentru }n\neq 1\mbox{)}\,\!$

$\int\frac{\sin^n cx\;dx}{\cos cx} = -\frac{\sin^{n-1} cx}{c(n-1)} + \int\frac{\sin^{n-2} cx\;dx}{\cos cx} \qquad\mbox{(pentru }n\neq 1\mbox{)}\,\!$

$\int\frac{\sin^n cx\;dx}{\cos^m cx} = \frac{\sin^{n+1} cx}{c(m-1)\cos^{m-1} cx}-\frac{n-m+2}{m-1}\int\frac{\sin^n cx\;dx}{\cos^{m-2} cx} \qquad\mbox{(pentru }m\neq 1\mbox{)}\,\!$

also: $\int\frac{\sin^n cx\;dx}{\cos^m cx} = -\frac{\sin^{n-1} cx}{c(n-m)\cos^{m-1} cx}+\frac{n-1}{n-m}\int\frac{\sin^{n-2} cx\;dx}{\cos^m cx} \qquad\mbox{(pentru }m\neq n\mbox{)}\,\!$

also: $\int\frac{\sin^n cx\;dx}{\cos^m cx} = \frac{\sin^{n-1} cx}{c(m-1)\cos^{m-1} cx}-\frac{n-1}{n-1}\int\frac{\sin^{n-1} cx\;dx}{\cos^{m-2} cx} \qquad\mbox{(pentru }m\neq 1\mbox{)}\,\!$

$\int\frac{\cos cx\;dx}{\sin^n cx} = -\frac{1}{c(n-1)\sin^{n-1} cx} \qquad\mbox{(pentru }n\neq 1\mbox{)}\,\!$

$\int\frac{\cos^2 cx\;dx}{\sin cx} = \frac{1}{c}\left(\cos cx+\ln\left|\tan\frac{cx}{2}\right|\right)$

$\int\frac{\cos^2 cx\;dx}{\sin^n cx} = -\frac{1}{n-1}\left(\frac{\cos cx}{c\sin^{n-1} cx)}+\int\frac{dx}{\sin^{n-2} cx}\right) \qquad\mbox{(pentru }n\neq 1\mbox{)}$

$\int\frac{\cos^n cx\;dx}{\sin^m cx} = -\frac{\cos^{n+1} cx}{c(m-1)\sin^{m-1} cx} - \frac{n-m-2}{m-1}\int\frac{cos^n cx\;dx}{\sin^{m-2} cx} \qquad\mbox{(pentru }m\neq 1\mbox{)}\,\!$

also: $\int\frac{\cos^n cx\;dx}{\sin^m cx} = \frac{\cos^{n-1} cx}{c(n-m)\sin^{m-1} cx} + \frac{n-1}{n-m}\int\frac{cos^{n-2} cx\;dx}{\sin^m cx} \qquad\mbox{(pentru }m\neq n\mbox{)}\,\!$

also: $\int\frac{\cos^n cx\;dx}{\sin^m cx} = -\frac{\cos^{n-1} cx}{c(m-1)\sin^{m-1} cx} - \frac{n-1}{m-1}\int\frac{cos^{n-2} cx\;dx}{\sin^{m-2} cx} \qquad\mbox{(pentru }m\neq 1\mbox{)}\,\!$

INTEGRALE de funcţii trigonometrice ce conţin atât sin cât şi tan

$\int \sin cx \tan cx\;dx = \frac{1}{c}(\ln|\sec cx + \tan cx| - \sin cx)\,\!$

$\int\frac{\tan^n cx\;dx}{\sin^2 cx} = \frac{1}{c(n-1)}\tan^{n-1} (cx) \qquad\mbox{(pentru }n\neq 1\mbox{)}\,\!$

INTEGRALE de funcţii trigonometrice ce conţin atât cos cât şi tan

$\int\frac{\tan^n cx\;dx}{\cos^2 cx} = \frac{1}{c(n+1)}\tan^{n+1} cx \qquad\mbox{(pentru }n\neq -1\mbox{)}\,\!$

INTEGRALE de funcţii trigonometrice ce conţin atât sin cât şi cot

$\int\frac{\cot^n cx\;dx}{\sin^2 cx} = \frac{1}{c(n+1)}\cot^{n+1} cx \qquad\mbox{(pentru }n\neq -1\mbox{)}\,\!$

INTEGRALE de funcţii trigonometrice ce conţin atât cos cât şi cot

$\int\frac{\cot^n cx\;dx}{\cos^2 cx} = \frac{1}{c(1-n)}\tan^{1-n} cx \qquad\mbox{(pentru }n\neq 1\mbox{)}\,\!$

INTEGRALE de funcţii trigonometrice ce conţin atât tan cât şi cot

$\int \frac{\tan^m(cx)}{\cot^n(cx)}\;dx = \frac{1}{c(m+n-1)}\tan^{m+n-1}(cx) - \int \frac{\tan^{m-2}(cx)}{\cot^n(cx)}\;dx\qquad\mbox{(pentru }m + n \neq 1\mbox{)}\,\!$

INTEGRALE de funcţii trigonometrice cu limitele simetrice

$\int_{{-c}}^{{c}}\sin {x}\;dx = 0 \!$

FUNCŢII HIPERBOLICE

$\int \sinh x \, dx = \cosh x + C$

$\int \cosh x \, dx = \sinh x + C$

$\int \tanh x \, dx = \ln |\cosh x| + C$

$\int \mbox{csch}\,x \, dx = \ln\left| \tanh {x \over2}\right| + C$

$\int \mbox{sech}\,x \, dx = \arctan(\sinh x) + C$

$\int \coth x \, dx = \ln|\sinh x| + C$

INTEGRALE DEFINITE CARE NU AU PRIMITIVE IMEDIATE

Există câteva funcții ale căror primitive (sau anti-derivate) nu pot fi exprimate într-o formă fixă, imediat vizibilă. Oricum, valoarea integralelor definite pe anumite intervale poate fi calculată. Unele dintre cel mai utile se găsesc mai jos:

$\int_0^\infty{\sqrt{x}\,e^{-x}\,dx} = \frac{1}{2}\sqrt \pi$ (a se vedea şi Funcția gamma)

$\int_0^\infty{e^{-x^2}\,dx} = \frac{1}{2}\sqrt \pi$ (Integrala lui Gauss - Gaussian integral)

$\int_0^\infty{\frac{x}{e^x-1}\,dx} = \frac{\pi^2}{6}$ (a se vedea și Numărul lui Bernoulli - Bernoulli number)

$\int_0^\infty{\frac{x^3}{e^x-1}\,dx} = \frac{\pi^4}{15}$

$\int_0^\infty\frac{\sin(x)}{x}\,dx=\frac{\pi}{2}$

$\int_0^\infty x^{z-1}\,e^{-x}\,dx = \Gamma(z)$ (în care $Γ($ $z$ $)$ este Funcția gamma)

$\int_{-\infty}^\infty e^{-(ax^2+bx+c)}\,dx=\sqrt{\frac{\pi}{a}}e^\frac{b^2-4ac}{4a}$

$\int_{0}^{2 \pi} e^{x \cos \theta} d \theta = 2 \pi I_{0}(x)$ (în care $I$ $0$ $($ $x$ $)$ este funcția Bessel modificată de ordinul întâi)

$\int_{0}^{2 \pi} e^{x \cos \theta + y \sin \theta} d \theta = 2 \pi I_{0} \left( \sqrt{x^2 + y^2} \right)$

CALCULAREA INTEGRALELOR DEFINITE

O nouă formă a metodei prin epuizare (exhaustivă) (în engleză, the method of exhaustion), furnizează o formulă de evaluare a integralelor definite pentru orice funcție continuă, utilă și în cazul în care aceaste integrale nu au primitive imediate.

$\int\limits_a^b {f(x)dx = \left( {b - a} \right)} \sum\limits_{n = 1}^\infty {\sum\limits_{m = 1}^{2^n - 1} {\left( { - 1} \right)^{m + 1} } } 2^{ - n} f(a + m\left( {b - a} \right)/2^n ).$

$\int\frac{(mx+n) dx}{ax^2+bx+c} =$	$\frac{m}{2a}\ln\left\|ax^2+bx+c\right\|+\frac{2an-bm}{a\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}} \qquad\mbox{(pentru }4ac-b^2>0\mbox{)}$
	$\frac{m}{2a}\ln\left\|ax^2+bx+c\right\|+\frac{2an-bm}{a\sqrt{b^2-4ac}}\,\mathrm{artanh}\frac{2ax+b}{\sqrt{b^2-4ac}} \qquad\mbox{(pentru }4ac-b^2<0\mbox{)}$
	$\frac{m}{2a}\ln\left\|ax^2+bx+c\right\|-\frac{2an-bm}{a(2ax+b)} \,\,\,\,\,\,\,\,\,\, \qquad\mbox{(pentru }4ac-b^2=0\mbox{)}$