Интеграл табу

Интеграл табу — математик анализда дифференциал табу белән нигез операциясе.

Интеграл табу кагыйдәләре

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

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

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

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

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

Элементар фунцияләр интеграллары

Рациональ функцияләр

$~\int \!0\,dx=C$

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

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

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

Исбатлау

Уң якта дифференциал табабыз: ${d \over dx}\,\left({1 \over a}\,\operatorname {arctg} \,{\frac {x}{a}}+C\right)={d \over dx}\,\left({1 \over a}\,\operatorname {arctg} \,{\frac {x}{a}}\right)={\frac {1}{a}}\cdot {d \over d\left({x \over a}\right)}\left(\operatorname {arctg} {\frac {x}{a}}\right)\cdot {d \over dx}\left({x \over a}\right)={\frac {1}{a}}\cdot {\frac {1}{1+{\frac {x^{2}}{a^{2}}}}}\cdot {\frac {1}{a}}={\frac {1}{a^{2}\cdot {\frac {a^{2}+x^{2}}{a^{2}}}}}={\frac {1}{a^{2}+x^{2}}}$

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

Логарифмнар

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

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

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

Экспонентлар

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

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

Иррациональ функцияләр

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

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

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

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

Тригонометрик функцияләр

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Гиперболоид функцияләр

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

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

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

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

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

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

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

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

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

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

Исбатлау

Исбатлау $\int \operatorname {sech} \,x\,dx=\arctan(\operatorname {sh} \,x)+C$ тикшерү белән башкарыйк: $\left(\operatorname {arctg} (\operatorname {sh} \,x\right)+C)'={\frac {\operatorname {ch} \,x}{\operatorname {sh} ^{2}\,x+1}}={\frac {\operatorname {ch} \,x}{\operatorname {ch} ^{2}\,x-1+1}}={\frac {1}{\operatorname {ch} \,x}}=\operatorname {sech} \,x$. Исбатлау $\int \operatorname {sech} \,x\,dx=2\operatorname {arctg} (e^{x})+C$ тикшерү белән башкарыйк: $\left(2\operatorname {arctg} (e^{x})+C\right)'={\frac {2e^{x}}{e^{2x}+1}}={\frac {2}{e^{x}+e^{-x}}}={\frac {1}{\operatorname {ch} \,x}}=\operatorname {sech} \,x$. Исбатлау $\int \operatorname {sech} \,x\,dx=2\,\operatorname {arctg} \,\left(\operatorname {th} \,{\frac {x}{2}}\right)+C$ тикшерәбез: $\left(2\,\operatorname {arctg} \,\left(\operatorname {th} \,{\frac {x}{2}}\right)+C\right)'={\frac {\operatorname {sech} ^{2}{\frac {x}{2}}}{\operatorname {th} ^{2}{\frac {x}{2}}+1}}={\frac {1}{\operatorname {sh} ^{2}{\frac {x}{2}}+\operatorname {ch} ^{2}{\frac {x}{2}}}}={\frac {1}{\operatorname {ch} \,x}}=\operatorname {sech} \,x$.