List of integrals of elementary functions

Integration is one of two basic operations in mathematical analysis . Unlike the differentiation operation, the integral of an elementary function may not be an elementary function. For example, it follows from the Liouville theorem that the integral of $e^{x^{2}}$ ${\ displaystyle e ^ {x ^ {2}}}$ $e ^ {x ^ 2}$ not an elementary function. Tables of the famous antiderivatives are often very useful, although now they are losing their relevance with the advent of computer algebra systems. This page provides a list of the most common antiderivatives.

$C$ ${\ displaystyle C}$ $C$ used as an arbitrary integration constant, which can be determined if the value of the integral at any point is known. Each function has an infinite number of antiderivatives.

Content

1 Function Integration Rules
2 Integrals of elementary functions
- 2.1 rational functions
- 2.2 Logarithms
- 2.3 Exponential Functions
- 2.4 Irrational functions
- 2.5 Trigonometric Functions
- 2.6 Hyperbolic Functions
3 Special functions
4 notes
5 Bibliography

Function Integration Rules

\int cf(x)\,dx=c\int f(x)\,dx

{\ displaystyle \ int cf (x) \, dx = c \ int f (x) \, dx}

\int cf(x)\,dx = c\int f(x)\,dx

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

{\ displaystyle \ int [f (x) + g (x)] \, dx = \ int f (x) \, dx + \ int g (x) \, dx}

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

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

{\ displaystyle \ int [f (x) -g (x)] \, dx = \ int f (x) \, dx- \ int g (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)\,df(x)

{\ displaystyle \ int f (x) g (x) \, dx = f (x) \ int g (x) \, dx- \ int \ left (\ int g (x) \, dx \ right) \, df (x)}

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

\int f(ax+b)\,dx={1 \over a}F(ax+b)\,+C

{\ displaystyle \ int f (ax + b) \, dx = {1 \ over a} F (ax + b) \, + C}

\int f(ax+b)\,dx = {1 \over a} F(ax+b)\,+C

Integrals of elementary functions

Rational Functions

\int \!0\,dx=C

{\ displaystyle \ int \! 0 \, dx = C}

\int \!0\,dx=C

(the antiderivative of zero is a constant; within any integration limits, the integral of zero is zero)

\int \!a\,dx=ax+C

{\ displaystyle \ int \! a \, dx = ax + C}

\int \!a\,dx=ax+C

\int \!x^{n}\,dx={\begin{cases}{\frac {x^{n+1}}{n+1}}+C,&n\neq -1\\\ln \left|x\right|+C,&n=-1\end{cases}}

{\ displaystyle \ int \! x ^ {n} \, dx = {\ begin {cases} {\ frac {x ^ {n + 1}} {n + 1}} + C, & n \ neq -1 \\ \ ln \ left | x \ right | + C, & n = -1 \ end {cases}}}

\int \!x^{n}\,dx={\begin{cases}{\frac {x^{n+1}}{n+1}}+C,&n\neq -1\\\ln \left|x\right|+C,&n=-1\end{cases}}

\int \!{dx \over {a^{2}+x^{2}}}={1 \over a}\,\operatorname {arctg} \,{\frac {x}{a}}+C=-{1 \over a}\,\operatorname {arcctg} \,{\frac {x}{a}}+C

{\ displaystyle \ int \! {dx \ over {a ^ {2} + x ^ {2}}} = {1 \ over a} \, \ operatorname {arctg} \, {\ frac {x} {a} } + C = - {1 \ over a} \, \ operatorname {arcctg} \, {\ frac {x} {a}} + C}

\int\!{dx \over {a^2+x^2}} = {1 \over a}\,\operatorname{arctg}\,\frac{x}{a} + C = - {1 \over a}\,\operatorname{arcctg}\,\frac{x}{a} + C

Evidence

Make a replacement $x=a\operatorname {tg} t$ ${\ displaystyle x = a \ operatorname {tg} t}$ we get

$\int \!{dx \over {a^{2}+x^{2}}}=\int \!{d(a\operatorname {tg} t) \over a^{2}+(a\operatorname {tg} t)^{2}}={1 \over a}\int \!{\cos ^{2}t \over \cos ^{2}t}dt={t \over a}+C={1 \over a}\operatorname {arctg} {x \over a}+C.$ ${\ displaystyle \ int \! {dx \ over {a ^ {2} + x ^ {2}}} = \ int \! {d (a \ operatorname {tg} t) \ over a ^ {2} + ( a \ operatorname {tg} t) ^ {2}} = {1 \ over a} \ int \! {\ cos ^ {2} t \ over \ cos ^ {2} t} dt = {t \ over a} + C = {1 \ over a} \ operatorname {arctg} {x \ over a} + C.}$

\int \!{dx \over {x^{2}-a^{2}}}={1 \over 2a}\ln \left|{x-a \over {x+a}}\right|+C

{\ displaystyle \ int \! {dx \ over {x ^ {2} -a ^ {2}}} = {1 \ over 2a} \ ln \ left | {xa \ over {x + a}} \ right | + C}

("High logarithm")

Logarithms

\int \!\ln {x}\,dx=x\ln {x}-x+C

{\ displaystyle \ int \! \ ln {x} \, dx = x \ ln {x} -x + C}

\int {\frac {dx}{x\ln x}}=\ln |\ln x|+C

{\ displaystyle \ int {\ frac {dx} {x \ ln x}} = \ ln | \ ln x | + C}

\int \!\log _{b}{x}\,dx=x\log _{b}{x}-x\log _{b}{e}+C=x{\frac {\ln {x}-1}{\ln b}}+C

{\ displaystyle \ int \! \ log _ {b} {x} \, dx = x \ log _ {b} {x} -x \ log _ {b} {e} + C = x {\ frac {\ ln {x} -1} {\ ln b}} + C}

Exponential Functions

\int \!e^{x}\,dx=e^{x}+C

{\ displaystyle \ int \! e ^ {x} \, dx = e ^ {x} + C}

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

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

Irrational Functions

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

{\ displaystyle \ 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

{\ displaystyle \ 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}\,\operatorname {arcsec} \,{|x| \over a}+C

{\ displaystyle \ int \! {dx \ over x {\ sqrt {x ^ {2} -a ^ {2}}}} = {1 \ over a} \, \ operatorname {arcsec} \, {| x |

\ over a} + C}

\int \!{dx \over {\sqrt {x^{2}+a}}}=\ln \left|{x+{\sqrt {x^{2}+a}}}\right|+C

{\ displaystyle \ int \! {dx \ over {\ sqrt {x ^ {2} + a}}} = \ ln \ left | {x + {\ sqrt {x ^ {2} + a}}} \ right | + C}

("Long logarithm")

\int \!{\sqrt {x^{2}+a}}\,dx={1 \over 2}({x}{\sqrt {x^{2}+a}}+{a}\ln |x+{\sqrt {x^{2}+a}}|)+C

{\ displaystyle \ int \! {\ sqrt {x ^ {2} + a}} \, dx = {1 \ over 2} ({x} {\ sqrt {x ^ {2} + a}} + {a } \ ln | x + {\ sqrt {x ^ {2} + a}} |) + C}

Evidence

Let be $a<0$ ${\ displaystyle a <0}$ , suppose also that $x\geq 0$ ${\ displaystyle x \ geq 0}$ . We use hyperbolic functions , make a replacement $x={\sqrt {-a}}\operatorname {ch} t,t\geq 0$ ${\ displaystyle x = {\ sqrt {-a}} \ operatorname {ch} t, t \ geq 0}$

${\begin{aligned}\int \!{\sqrt {x^{2}+a}}dx&=\int {\sqrt {({\sqrt {-a}}\operatorname {ch} t)^{2}+a}}d({\sqrt {-a}}\operatorname {ch} t)=-a\int {\sqrt {\operatorname {ch} ^{2}t-1}}\operatorname {sh} tdt\\&=-a\int \operatorname {sh} ^{2}tdt=-a\int {\operatorname {ch} 2t-1 \over 2}dt={-a \over 2}\left({\operatorname {sh} 2t \over 2}-t\right)+C_{1}\\&={-a \over 2}(\operatorname {sh} t\operatorname {ch} t-t)+C_{1}\end{aligned}}$ ${\ displaystyle {\ begin {aligned} \ int \! {\ sqrt {x ^ {2} + a}} dx & = \ int {\ sqrt {({\ sqrt {-a}} \ operatorname {ch} t) ^ {2} + a}} d ({\ sqrt {-a}} \ operatorname {ch} t) = - a \ int {\ sqrt {\ operatorname {ch} ^ {2} t-1}} \ operatorname {sh} tdt \\ & = - a \ int \ operatorname {sh} ^ {2} tdt = -a \ int {\ operatorname {ch} 2t-1 \ over 2} dt = {- a \ over 2} \ left ({\ operatorname {sh} 2t \ over 2} -t \ right) + C_ {1} \\ & = {- a \ over 2} (\ operatorname {sh} t \ operatorname {ch} tt) + C_ {1} \ end {aligned}}}$

But

$\operatorname {sh} t={\sqrt {\operatorname {ch} ^{2}-1}}={\sqrt {{x^{2} \over -a}-1}}={{\sqrt {x^{2}+a}} \over {\sqrt {-a}}},$ ${\ displaystyle \ operatorname {sh} t = {\ sqrt {\ operatorname {ch} ^ {2} -1}} = {\ sqrt {{x ^ {2} \ over -a} -1}} = {{ \ sqrt {x ^ {2} + a}} \ over {\ sqrt {-a}}},}$ $\operatorname {sh} t\operatorname {ch} t=x{{\sqrt {x^{2}+a}} \over -a},$ ${\ displaystyle \ operatorname {sh} t \ operatorname {ch} t = x {{\ sqrt {x ^ {2} + a}} \ over -a},}$ $e^{t}=\operatorname {sh} t+\operatorname {ch} t={x+{\sqrt {x^{2}+a}} \over {\sqrt {-a}}}.$ ${\ displaystyle e ^ {t} = \ operatorname {sh} t + \ operatorname {ch} t = {x + {\ sqrt {x ^ {2} + a}} \ over {\ sqrt {-a}}}.}$

therefore

$t=\ln {x+{\sqrt {x^{2}+a}} \over {\sqrt {-a}}}.$ ${\ displaystyle t = \ ln {x + {\ sqrt {x ^ {2} + a}} \ over {\ sqrt {-a}}}.}$

Hence, including the logarithm of the denominator of the last fraction in the constant C, we obtain

$\int \!{\sqrt {x^{2}+a}}\,dx={x \over 2}{\sqrt {x^{2}+a}}+{a \over 2}\ln |x+{\sqrt {x^{2}+a}}|+C$ ${\ displaystyle \ int \! {\ sqrt {x ^ {2} + a}} \, dx = {x \ over 2} {\ sqrt {x ^ {2} + a}} + {a \ over 2} \ ln | x + {\ sqrt {x ^ {2} + a}} | + C}$

If $x<0$ ${\ displaystyle x <0}$ then by replacing $x=-t,t>0$ ${\ displaystyle x = -t, t> 0}$ we reduce the integral to the case already considered. If $a>0$ ${\ displaystyle a> 0}$ then make a replacement $x={\sqrt {a}}\operatorname {sh} t$ ${\ displaystyle x = {\ sqrt {a}} \ operatorname {sh} t}$ and we carry out arguments similar to the case considered ^[1] .

Trigonometric Functions

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

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

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

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

\int \!\operatorname {tg} \,{x}\,dx=-\ln {\left|\cos {x}\right|}+C

{\ displaystyle \ int \! \ operatorname {tg} \, {x} \, dx = - \ ln {\ left | \ cos {x} \ right |} + C}

Evidence

$\int \!\operatorname {tg} \,{x}\,dx=\int {\frac {\sin x}{\cos x}}dx=-\int {\frac {d(\cos x)}{\cos x}}=-\ln |\cos x|+C$ ${\ displaystyle \ int \! \ operatorname {tg} \, {x} \, dx = \ int {\ frac {\ sin x} {\ cos x}} dx = - \ int {\ frac {d (\ cos x)} {\ cos x}} = - \ ln | \ cos x | + C}$

\int \!\operatorname {ctg} \,{x}\,dx=\ln {\left|\sin {x}\right|}+C

{\ displaystyle \ int \! \ operatorname {ctg} \, {x} \, dx = \ ln {\ left | \ sin {x} \ right |} + C}

Evidence

$\int \!\operatorname {ctg} \,{x}\,dx=\int {\frac {\cos x}{\sin x}}dx=\int {\frac {d(\sin x)}{\sin x}}=\ln |\sin x|+C$ ${\ displaystyle \ int \! \ operatorname {ctg} \, {x} \, dx = \ int {\ frac {\ cos x} {\ sin x}} dx = \ int {\ frac {d (\ sin x )} {\ sin x}} = \ ln | \ sin x | + C}$

\int \!\sec {x}\,dx=\ln {\left|\sec {x}+\operatorname {tg} \,{x}\right|}+C

{\ displaystyle \ int \! \ sec {x} \, dx = \ ln {\ left | \ sec {x} + \ operatorname {tg} \, {x} \ right |} + C}

\int \!\operatorname {cosec} {x}\,dx=-\ln {\left|\operatorname {cosec} {x}+\operatorname {ctg} \,{x}\right|}+C

{\ displaystyle \ int \! \ operatorname {cosec} {x} \, dx = - \ ln {\ left | \ operatorname {cosec} {x} + \ operatorname {ctg} \, {x} \ right |} + C}

\int \!\sec ^{2}x\,dx=\int \!{dx \over \cos ^{2}x}=\operatorname {tg} \,x+C

{\ displaystyle \ int \! \ sec ^ {2} x \, dx = \ int \! {dx \ over \ cos ^ {2} x} = \ operatorname {tg} \, x + C}

\int \!\operatorname {cosec} ^{2}x\,dx=\int \!{dx \over \sin ^{2}x}=-\operatorname {ctg} \,x+C

{\ displaystyle \ int \! \ operatorname {cosec} ^ {2} x \, dx = \ int \! {dx \ over \ sin ^ {2} x} = - \ operatorname {ctg} \, x + C}

\int \!\sec {x}\,\operatorname {tg} \,{x}\,dx=\sec {x}+C

{\ displaystyle \ int \! \ sec {x} \, \ operatorname {tg} \, {x} \, dx = \ sec {x} + C}

\int \!\operatorname {cosec} {x}\,\operatorname {ctg} \,{x}\,dx=-\operatorname {cosec} {x}+C

{\ displaystyle \ int \! \ operatorname {cosec} {x} \, \ operatorname {ctg} \, {x} \, dx = - \ operatorname {cosec} {x} + C}

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

{\ displaystyle \ 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

{\ displaystyle \ 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,n\in \mathbb {N} ,n\geqslant 2

{\ displaystyle \ int \! \ sin ^ {n} x \, dx = - {\ frac {\ sin ^ {n-1} {x} \ cos {x}} {n}} + {\ frac {n -1} {n}} \ int \! \ Sin ^ {n-2} {x} \, dx, n \ in \ mathbb {N}, n \ geqslant 2}

\int \!\cos ^{n}x\,dx={\frac {\cos ^{n-1}{x}\sin {x}}{n}}+{\frac {n-1}{n}}\int \!\cos ^{n-2}{x}\,dx,n\in \mathbb {N} ,n\geqslant 2

{\ displaystyle \ int \! \ cos ^ {n} x \, dx = {\ frac {\ cos ^ {n-1} {x} \ sin {x}} {n}} + {\ frac {n- 1} {n}} \ int \! \ Cos ^ {n-2} {x} \, dx, n \ in \ mathbb {N}, n \ geqslant 2}

\int \!\operatorname {arctg} \,{x}\,dx=x\,\operatorname {arctg} \,{x}-{\frac {1}{2}}\ln {\left(1+x^{2}\right)}+C

{\ displaystyle \ int \! \ operatorname {arctg} \, {x} \, dx = x \, \ operatorname {arctg} \, {x} - {\ frac {1} {2}} \ ln {\ left (1 + x ^ {2} \ right)} + C}

Hyperbolic Functions

\int \operatorname {sh} \,x\,dx=\operatorname {ch} \,x+C

{\ displaystyle \ int \ operatorname {sh} \, x \, dx = \ operatorname {ch} \, x + C}

\int \operatorname {ch} \,x\,dx=\operatorname {sh} \,x+C

{\ displaystyle \ int \ operatorname {ch} \, x \, dx = \ operatorname {sh} \, x + C}

\int {\frac {dx}{\operatorname {ch} ^{2}\,x}}=\operatorname {th} \,x+C

{\ displaystyle \ int {\ frac {dx} {\ operatorname {ch} ^ {2} \, x}} = \ operatorname {th} \, x + C}

\int {\frac {dx}{\operatorname {sh} ^{2}\,x}}=-\operatorname {cth} \,x+C

{\ displaystyle \ int {\ frac {dx} {\ operatorname {sh} ^ {2} \, x}} = - \ operatorname {cth} \, x + C}

\int \operatorname {th} \,x\,dx=\ln |\operatorname {ch} \,x|+C

{\ displaystyle \ int \ operatorname {th} \, x \, dx = \ ln | \ operatorname {ch} \, x | + C}

\int \operatorname {csch} \,x\,dx=\ln \left|\operatorname {th} \,{x \over 2}\right|+C

{\ displaystyle \ int \ operatorname {csch} \, x \, dx = \ ln \ left | \ operatorname {th} \, {x \ over 2} \ right | + C}

\int \operatorname {sech} \,x\,dx=\operatorname {arctg} \,\operatorname {sh} \,x+C

{\ displaystyle \ int \ operatorname {sech} \, x \, dx = \ operatorname {arctg} \, \ operatorname {sh} \, x + C}

also

\int \operatorname {sech} \,x\,dx=2\,\operatorname {arctg} \,(e^{x})+C

{\ displaystyle \ int \ operatorname {sech} \, x \, dx = 2 \, \ operatorname {arctg} \, (e ^ {x}) + C}

also

\int \operatorname {sech} \,x\,dx=2\,\operatorname {arctg} \,\left(\operatorname {th} \,{\frac {x}{2}}\right)+C

{\ displaystyle \ int \ operatorname {sech} \, x \, dx = 2 \, \ operatorname {arctg} \, \ left (\ operatorname {th} \, {\ frac {x} {2}} \ right) + C}

\int \operatorname {cth} \,x\,dx=\ln |\operatorname {sh} \,x|+C

{\ displaystyle \ int \ operatorname {cth} \, x \, dx = \ ln | \ operatorname {sh} \, x | + C}

Proof of

Proof of the formula $\int \operatorname {sech} \,x\,dx=\operatorname {arctg} \operatorname {sh} \,x+C$ ${\ displaystyle \ int \ operatorname {sech} \, x \, dx = \ operatorname {arctg} \ operatorname {sh} \, x + C}$ :

\int \operatorname {sech} \,x\,dx=\int {dx \over \operatorname {ch} x}=\int {\operatorname {ch} x \over \operatorname {ch} ^{2}x}dx=\int {d(\operatorname {sh} x) \over 1+\operatorname {sh} ^{2}x}=\operatorname {arctg} \operatorname {sh} x+C

{\ displaystyle \ int \ operatorname {sech} \, x \, dx = \ int {dx \ over \ operatorname {ch} x} = \ int {\ operatorname {ch} x \ over \ operatorname {ch} ^ {2 } x} dx = \ int {d (\ operatorname {sh} x) \ over 1+ \ operatorname {sh} ^ {2} x} = \ operatorname {arctg} \ operatorname {sh} x + C}

Proof of the formula $\int \operatorname {sech} \,x\,dx=2\operatorname {arctg} (e^{x})+C$ ${\ displaystyle \ int \ operatorname {sech} \, x \, dx = 2 \ operatorname {arctg} (e ^ {x}) + C}$ : $\int \operatorname {sech} \,x\,dx=\int {dx \over \operatorname {ch} x}=2\int {dx \over e^{x}+e^{-x}}=2\int {d{e^{x}} \over 1+e^{2x}}=2\operatorname {arctg} (e^{x})+C$ ${\ displaystyle \ int \ operatorname {sech} \, x \, dx = \ int {dx \ over \ operatorname {ch} x} = 2 \ int {dx \ over e ^ {x} + e ^ {- x} } = 2 \ int {d {e ^ {x}} \ over 1 + e ^ {2x}} = 2 \ operatorname {arctg} (e ^ {x}) + C}$ .

Proof of the formula $\int \operatorname {sech} \,x\,dx=2\,\operatorname {arctg} \,\left(\operatorname {th} \,{\frac {x}{2}}\right)+C$ ${\ displaystyle \ int \ operatorname {sech} \, x \, dx = 2 \, \ operatorname {arctg} \, \ left (\ operatorname {th} \, {\ frac {x} {2}} \ right) + C}$ :

{\begin{aligned}\int \operatorname {sech} \,x\,dx&=\int {1 \over \operatorname {ch} x}dx=\int {dx \over \operatorname {sh} ^{2}{x \over 2}+\operatorname {ch} ^{2}{x \over 2}}=2\int {d({x \over 2}) \over \operatorname {ch} ^{2}{x \over 2}(1+\operatorname {th} ^{2}{x \over 2})}\\&=2\int {d(\operatorname {th} {x \over 2}) \over 1+\operatorname {th} ^{2}{x \over 2}}=2\,\operatorname {arctg} \,\left(\operatorname {th} \,{\frac {x}{2}}\right)+C\end{aligned}}

{\ displaystyle {\ begin {aligned} \ int \ operatorname {sech} \, x \, dx & = \ int {1 \ over \ operatorname {ch} x} dx = \ int {dx \ over \ operatorname {sh} ^ {2} {x \ over 2} + \ operatorname {ch} ^ {2} {x \ over 2}} = 2 \ int {d ({x \ over 2}) \ over \ operatorname {ch} ^ {2 } {x \ over 2} (1+ \ operatorname {th} ^ {2} {x \ over 2})} \\ & = 2 \ int {d (\ operatorname {th} {x \ over 2}) \ over 1+ \ operatorname {th} ^ {2} {x \ over 2}} = 2 \, \ operatorname {arctg} \, \ left (\ operatorname {th} \, {\ frac {x} {2}} \ right) + C \ end {aligned}}}

Special Functions

\int \operatorname {Ci} (x)\,dx=x\operatorname {Ci} (x)-\sin x

{\ displaystyle \ int \ operatorname {Ci} (x) \, dx = x \ operatorname {Ci} (x) - \ sin x}

\int \operatorname {Si} (x)\,dx=x\operatorname {Si} (x)+\cos x

{\ displaystyle \ int \ operatorname {Si} (x) \, dx = x \ operatorname {Si} (x) + \ cos x}

\int \operatorname {Ei} (x)\,dx=x\operatorname {Ei} (x)-e^{x}

{\ displaystyle \ int \ operatorname {Ei} (x) \, dx = x \ operatorname {Ei} (x) -e ^ {x}}

\int \operatorname {li} (x)\,dx=x\operatorname {li} (x)-\operatorname {Ei} (2\ln x)

{\ displaystyle \ int \ operatorname {li} (x) \, dx = x \ operatorname {li} (x) - \ operatorname {Ei} (2 \ ln x)}

\int {\frac {\operatorname {li} (x)}{x}}\,dx=\ln x\,\operatorname {li} (x)-x

{\ displaystyle \ int {\ frac {\ operatorname {li} (x)} {x}} \, dx = \ ln x \, \ operatorname {li} (x) -x}

\int \operatorname {erf} (x)\,dx={\frac {e^{-x^{2}}}{\sqrt {\pi }}}+x\operatorname {erf} (x)

{\ displaystyle \ int \ operatorname {erf} (x) \, dx = {\ frac {e ^ {- x ^ {2}}} {\ sqrt {\ pi}}} + x \ operatorname {erf} (x )}

Notes

↑ Vinogradova I.A., Olehnik S.N., Sadovnichy V.A. Tasks and exercises in mathematical analysis. In 2 book Prince 1 / Ed. V.A. Gardener. - 2nd ed. - M .: Higher school , 2000. - S. 187. - ISBN 5-06-003768-1 .

Bibliography

Books

Gradshtein I. S. Ryzhik I. M. Tables of integrals, sums, series and products. - 4th ed. - M .: Nauka, 1963. - ISBN 0-12-294757-6 // EqWorld
Dvayt G. B. Tables of Integrals of St. Petersburg: Publishing House and Printing House of VNIIG im. B.V. Vedeneeva, 1995 .-- 176 p. - ISBN 5-85529-029-8 .
D. Zwillinger. CRC Standard Mathematical Tables and Formulae , 31st ed., 2002. ISBN 1-58488-291-3 .
M. Abramowitz and IA Stegun, eds. Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables , 1964. ISBN 0-486-61272-4
Korn G.A., Korn T.M. Math reference book for scientists and engineers . - M .: " Science ", 1974.

Integral Tables

Calculation of Integrals

The Integrator (at Wolfram Research )
Empire of Numbers

[1] Vinogradova I.A., Olehnik S.N., Sadovnichy V.A. Tasks and exercises in mathematical analysis. In 2 book Prince 1 / Ed. V.A. Gardener. - 2nd ed. - M .: Higher school , 2000. - S. 187. - ISBN 5-06-003768-1 .

List of integrals of elementary functions

Content

Function Integration Rules

Integrals of elementary functions

Rational Functions

Logarithms

Exponential Functions

Irrational Functions

Trigonometric Functions

Hyperbolic Functions

Special Functions

Notes

Bibliography

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