Интеграл табу

Интеграл табу — математик анализда дифференциал табу белән нигез операциясе.

\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

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 \!\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

\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

\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

\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 $\int \operatorname {sech} \,x\,dx=\operatorname {arctg} (\operatorname {sh} \,x)+C$ 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: $\left(2\,\operatorname {arctg} \,\left(\operatorname {th} \,{\frac {x}{2}}\right)+C\right)'={\frac {\operatorname {sech} ^{2}{\frac {x}{2}}}{\operatorname {th} ^{2}{\frac {x}{2}}+1}}={\frac {1}{\operatorname {sh} ^{2}{\frac {x}{2}}+\operatorname {ch} ^{2}{\frac {x}{2}}}}={\frac {1}{\operatorname {ch} \,x}}=\operatorname {sech} \,x$ .