查看“︁微积分学/不定积分/部分积分法”︁的源代码

{{wikipedia|部分分式积分法}}
==例題==
<math>
\begin{align}
&
1.\;\int\frac{x^3+4x+3}{x^4+5x^2+4}dx
\\&
=\int\frac{x^3+4x+3}{\left(x^2+1\right)\left(x^2+4\right)}dx
\\&
=\int\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+4}dx
\\&
\frac{x^3+4x+3}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+4}
\\&
\frac{x^3+4x+3}{\left(x^2+4\right)}=Ax+B+\frac{Cx+D}{x^2+4}\left(x^2+1\right)
\\&
x=i
\\&
\quad\frac{-i+4i+3}3=i+1=Ai+B
\\&
\qquad\qquad\qquad\qquad\qquad\quad A=1\qquad B=1
\\&
\frac{x^4+4x^2+3x}{x^4+5x^2+4}=\frac{x^2+x}{x^2+1}+\frac{Cx^2+Dx}{x^2+4}
\\&
x\to\infty
\\&
\quad1=1+C\qquad\qquad C=0
\\&
x=0
\\&
\quad\frac34=1+\frac D4\qquad\quad D=-1
\\&
\int\frac{x^3+4x+3}{\left(x^2+1\right)\left(x^2+4\right)}dx
\\&
=\int\frac{x+1}{x^2+1}+\frac{-1}{x^2+4}dx
\\&
=\int\frac x{x^2+1}dx+\int\frac1{x^2+1}dx+\int\frac{-1}{x^2+4}dx
\\&
=\frac{\ln\left(x^2+1\right)}2+\tan^{-1}x-\frac12\tan^{-1}\frac x2
\\&
\\&
2.\;\int\frac1{x+x\sqrt x}dx
\\&
=2\int\frac1{\sqrt x+x}d\sqrt x=2\int\left(\frac A\sqrt x+\frac B{\sqrt x+1}\right)d\sqrt x
\\&
\quad A=1\quad B=-1
\\&
=2\int\left(\frac1\sqrt x+\frac{-1}{\sqrt x+1}\right)d\sqrt x
\\&
=2\left(\int\frac1\sqrt xd\sqrt x-\int\frac1{\sqrt x+1}d\sqrt x\right)
\\&
=2\left(\ln\sqrt x-\ln\left(\sqrt x+1\right)\right)
\\&
=2\ln\frac\sqrt x{\sqrt x+1}
\end{align}
</math>