Indefinite Integrals

∫xax+bdx=xa−ba2ln⁡∣ax+b∣\int\frac{x}{ax + b} dx = \frac x a - \frac b {a^2}\ln|ax + b|
∫1x(ax+b)dx=−1bln⁡∣ax+bx∣\int\frac{1}{x(ax + b)}dx = -\frac 1 b\ln\left|\frac{ax + b} x\right|
∫ax+bfx+gdx=axf+bf−agf2ln⁡∣fx+g∣\int\frac{ax + b}{fx + g}dx = \frac{ax} f + \frac{bf - ag}{f^2}\ln|fx + g|
∫x(ax+b)(fx+g)dx=1bf−ag[baln⁡∣ax+b∣−gfln⁡∣fx+g∣]\int \frac x{(ax + b)(fx + g)}dx = \frac 1{bf - ag}\left[\frac b a \ln|ax + b| - \frac g f\ln|fx + g|\right]
Definition:χ={24ac−b2arctan⁡2ax+b4ac−b2if  4ac>b21a(p−q)ln⁡∣x−px−q∣if  4ac<b2\textup{Definition:}\quad \chi = \begin{cases} \frac 2{\sqrt{4ac - b^2}}\arctan\frac{2ax + b}{\sqrt{4ac - b^2}} & \textup{if} \; 4ac > b^2 \\ \frac 1{a(p - q)}\ln\left|\frac{x - p}{x - q}\right| & \textup{if} \; 4ac < b^2 \end{cases}

Where

pp

and

qq

are the roots of

ax2+bx+c=0ax^2 + bx + c = 0

.

∫1ax2+bx+cdx=χ\int \frac 1{ax^2 + bx + c}dx = \chi
∫xax2+bx+cdx=12aln⁡∣ax2+bx+c∣−b2aχ\int \frac x{ax^2 + bx + c}dx = \frac 1{2a}\ln\left|ax^2 + bx + c\right| - \frac b{2a}\chi
∫x2ax2+bx+cdx=xa−b2a2ln⁡∣ax2+bx+c∣+b2−2ac2a2χ\int\frac{x^2}{ax^2 + bx + c}dx = \frac xa - \frac b{2a^2}\ln\left|ax^2 + bx + c\right| + \frac{b^2 - 2ac}{2a^2}\chi
∫1(ax2+bx+c)2dx=2ax+b(4ac−b2)(ax2+bx+c)+2a(4ac−b2)χ\int\frac 1{(ax^2 + bx + c)^2}dx = \frac{2ax + b}{(4ac - b^2)(ax^2 + bx + c)} + \frac{2a}{(4ac - b^2)}\chi
∫x(ax2+bx+c)2dx=−bx+2c(4ac−b2)(ax2+bx+c)−b(4ac−b2)χ\int\frac x{(ax^2 + bx + c)^2}dx = -\frac{bx+2c}{(4ac - b^2)(ax^2 + bx + c)} - \frac b{(4ac - b^2)}\chi
∫ax+bdx=23a(ax+b)3/2\int\sqrt{ax + b}dx = \frac 2{3a}(ax + b)^{3/2}
∫xax+bdx=2(3ax−2b)15a2(ax+b)3/2\int x\sqrt{ax + b}dx = \frac{2(3ax - 2b)}{15a^2}(ax + b)^{3/2}
∫1ax+bdx=2ax+ba\int\frac 1{\sqrt{ax + b}}dx = \frac{2\sqrt{ax + b}}a
∫xax+bdx=2(ax−2b)3a2ax+b\int\frac x{\sqrt{ax + b}}dx = \frac{2(ax - 2b)}{3a^2}\sqrt{ax + b}
∫a2−x2dx=x2a2−x2+a22arcsin⁡xa\int\sqrt{a^2 - x^2}dx = \frac x 2\sqrt{a^2 - x^2} + \frac {a^2}2 \arcsin\frac x a
∫xa2−x2dx=−13(a2−x2)3/2\int x\sqrt{a^2 - x^2}dx = -\frac 1 3(a^2 - x^2)^{3/2}
∫1a2−x2dx=arcsin⁡xa\int\frac 1{\sqrt{a^2 - x^2}}dx = \arcsin\frac x a
∫xa2−x2dx=−a2−x2\int\frac x{\sqrt{a^2 - x^2}}dx = -\sqrt{a^2 - x^2}
∫1xa2−x2dx=−1aln⁡∣a+a2−x2x∣\int\frac 1{x\sqrt{a^2 - x^2}}dx = -\frac 1 a\ln\left|\frac{a + \sqrt{a^2 - x^2}}x\right|
∫x2−a2dx=x2x2−a2−a22ln⁡∣x+x2−a2∣\int\sqrt{x^2 - a^2}dx =\frac x 2\sqrt{x^2 - a^2} - \frac{a^2}2\ln\left|x + \sqrt{x^2 - a^2}\right|
∫1x2−a2dx=ln⁡∣x+x2−a2∣\int\frac 1{\sqrt{x^2 - a^2}}dx = \ln\left|x + \sqrt{x^2 - a^2}\right|
∫x2+a2dx=x2x2+a2+a22ln⁡∣x+x2+a2∣\int\sqrt{x^2 + a^2}dx =\frac x 2\sqrt{x^2 + a^2} + \frac{a^2}2\ln\left|x + \sqrt{x^2 + a^2}\right|
∫1x2+a2dx=ln⁡∣x+x2+a2∣\int\frac 1{\sqrt{x^2 + a^2}}dx = \ln\left|x + \sqrt{x^2 + a^2}\right|
∫1sin⁡axdx=1aln⁡∣tan⁡ax2∣\int\frac 1{\sin ax}dx = \frac 1 a \ln\left|\tan\frac{ax}2\right|
∫1cos⁡axdx=1aln⁡∣tan⁡(ax2+π4)∣\int\frac 1{\cos ax}dx = \frac 1 a\ln\left|\tan\left(\frac{ax}2 + \frac \pi 4\right)\right|
∫tan⁡ax dx=−1aln⁡∣cos⁡ax∣\int\tan ax \:dx =-\frac 1 a\ln|\cos ax|
∫cot⁡ax dx=1aln⁡∣sin⁡ax∣\int\cot ax \: dx = \frac 1 a \ln|\sin ax|
∫xsin⁡x dx=sin⁡x−xcos⁡x\int x\sin x \: dx = \sin x - x\cos x
∫xsin⁡2x dx=x24−xsin⁡2x4−cos⁡2x8\int x\sin^2 x \: dx = \frac{x^2}4 -\frac{x\sin 2x}4 - \frac{\cos 2x}8
∫sin⁡2x dx=12(x−sin⁡xcos⁡x)\int\sin^2 x\: dx = \frac 1 2(x - \sin x\cos x)
∫ln⁡x dx=xln⁡∣x∣−x\int\ln x \: dx = x \ln|x| - x
∫xeax dx=eaxa2(ax−1)\int xe^{ax} \: dx =\frac{e^{ax}}{a^2}(ax - 1)
∫eaxsin⁡bx dx=eaxa2+b2(asin⁡bx−bcos⁡bx)\int e^{ax} \sin bx\: dx = \frac{e^{ax}}{a^2 + b^2}(a\sin bx - b\cos bx)

Definite Integrals

∫0∞x2n⋅e−ax2 dx=(2n−1)!!2(2a)nπa\int_0^\infty x^{2n}\cdot e^{-ax^2} \: dx = \frac{(2n - 1)!!}{2(2a)^n}\sqrt\frac \pi a

Where

a>0a > 0

,

n!=n !=

negative integer.

∫0∞x2n+1⋅e−ax2 dx=n!2an+1\int_0^\infty x^{2n + 1}\cdot e^{-ax^2} \: dx = \frac{n!}{2a^{n+1}}
∫0∞xk⋅e−ax dx=Γ(k+1)⋅a−(k+1)\int_0^\infty x^k\cdot e^{-ax}\: dx = \Gamma(k + 1)\cdot a^{-(k + 1)}
Ik=∫0∞xkex−1 dx=Γ(k+1)⋅ζ(k+1)I_k = \int_0^\infty \frac{x^k}{e^x - 1}\: dx = \Gamma(k + 1)\cdot \zeta(k + 1)
Jk=∫0∞xkex+1 dx=(1−2−k)⋅IkJ_k = \int_0^\infty \frac{x^k}{e^x + 1}\: dx = (1 - 2^{-k})\cdot I_k
ζ(k)=∑n=1∞1nk\zeta(k) = \sum_{n = 1}^\infty \frac 1 {n^k}
∫0π/2sin⁡2a+1x⋅cos⁡2b+1x dx=Γ(a+1)⋅Γ(b+1)2Γ(a+b+2)\int_0^{\pi/2} \sin^{2a + 1}x\cdot\cos^{2b + 1}x \: dx = \frac{\Gamma(a + 1)\cdot\Gamma(b + 1)}{2\Gamma(a + b + 2)}

Stirling's approximation

N!≈2πN(Ne)NN≫1N! \approx \sqrt{2\pi N}\left(\frac N e\right)^N \quad N \gg 1

Error Function

erf(x)=2π∫0xe−ξ2 dξ\textup{erf}(x) = \frac 2{\sqrt\pi}\int_0^x e^{-\xi^2} \: d\xi
erfc(x)=1−erf(x)=2π∫x∞e−ξ2 dξ\textup{erfc}(x) = 1 - \textup{erf}(x) = \frac 2{\sqrt\pi}\int_x^\infty e^{-\xi^2} \: d\xi
erf(∞)=1\textup{erf}(\infty) = 1