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Integrate the rational functions$\frac{x}{(x^2+1)(x-1)}$

Toolbox:
• $(i)\;$Form of the rational function$\frac{px+q}{(x+a)(x^2+b)}$
• $\;$Form of the partial function$\frac{A}{x+a}+\frac{Bx}{x^2+b}$
• $(ii)\;\int\frac{dx}{(x+a)}=log|x+a|+c.$
• $(iii)\int\frac{dx}{x^2+1}=\tan^{-1}x+c.$
Given $I=\int\frac{x}{(x^2+1)(x-1)}dx.$

$\frac{x}{(x^2+1)(x-1)}=\frac{Ax+B}{x^2+1}+\frac{C}{x-1}.$

$\Rightarrow x=(Ax+B)(x-1)+C(x^2+1).$

equating the coefficients of $x^2$,

0=A+C.------(1)

equating the coefficients of x,

1=-A+B------(2)

Equating the constant terms,

0=-B+C-------(3)

Now solving the three equations will give A,B,C eliminate A by adding equ(1) and equ(2)
A+C=0
-A+B=1
_______________
B+C=1--------(4)
-B+C=0
B+C=1
_______________
2C=1$\Rightarrow C=\frac{1}{2}$.

substitute for C in equ(3),we get

$B=\frac{1}{2}.$

substitute for B in equ(2),we get

$-A+\frac{1}{2}=1\Rightarrow A=\frac{-1}{2}.$

$A=\frac{-1}{2},B=\frac{1}{2},C=\frac{1}{2}.$

$I=\int\frac{-1/2x+1/2}{(x^2+1)}dx+\int\frac{1/2}{(x-1)}dx.$

$\;\;\;=\frac{-1}{2}\int\frac{x}{x^2+1}dx+\frac{1}{2}\int\frac{dx}{x^2+1}+\frac{1}{2}\int\frac{dx}{x-1}.$

On integrating we get,

Let $x^2+1=t.$

Therefore $2xdx=dt\Rightarrow xdx=\frac{dt}{2}.$

Substituting this we get,

$I=\frac{-1}{2}\int\frac{dt/2}{t}+\frac{1}{2}\int{dx}{x^2+1}+\frac{1}{2}\int\frac{dx}{x-1}.$

$\;\;\;=\frac{-1}{4}\frac{dt}{t}+\frac{1}{2}\int\frac{dx}{x^2+1}+\frac{1}{2}\int\frac{dx}{x-1}.$

$\;\;\;=\frac{-1}{4}|log t|+\frac{1}{2}\tan^{-1}x+\frac{1}{2}log|x-1|.$

Substituting back for t we get,

$\;\;\;=\frac{-1}{4}log|x^2+1|+\frac{1}{2}\tan^{-1}x+\frac{1}{2}log|x-1|+c.$

edited Feb 5, 2013