# Integrate the rational functions$\frac{2x}{(x^2+1)(x^2+3)}$

Toolbox:
• $(i)\;$Form of the rational function$\frac{px+q}{(x^2+a)(x^2+b)}$
• $\;$Form of the partial function$\frac{Ax+B}{(x^2+a)}+\frac{Cx+D}{(x^2+b)}$
• $(ii)\int\frac{dx}{x}=log|x|+c.$
Given$I=\int \frac{2x}{(x^2+1)(x^2+3)}$

Resolving into partial functions

Let $\frac{2x}{(x^2+1)(x^2+3)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+3}.$

$\Rightarrow 2x=(Ax+B)(x^2+3)+(Cx+D)(x^2+1).$

Equating the coefficients of $x^3$.

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

Equating the coefficients of $x^2$,

0=B+D----(2)

Equating the coefficients of $x$,

2=3A+C----(3)

Equating the constant terms

0=3B+D----(4)

Consider equ(1) and equ(3)

On subtracting them we get,

A+C=0
3A+C=2
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-2A=-2$\Rightarrow A=1.$
Substitute for A in eq(1)

A+C=0.

1+C=0$\Rightarrow C=-1.$

Consider eq(2) and eq(4),

On subtracting them we get,

B+D=0.

3B+D=0.

since both the equations are equated to 0,B and D are 0.

Hence A=1 and C=-1.

Now substituting for A and C in I we get,

$\frac{2x}{(x^2+1)(x^2+3)}=\frac{x}{x^2+1}-\frac{x}{x^2+3}.$

For integrating we can use the substitution method,

$I=\int\frac{x}{x^2+1}dx-\int\frac{x}{x^2+3}dx.$

Let $x^2=t.$

On differentiating we get,

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

Similarly let $x^2+3=u$,on differentiating we get,

2xdx=du$\Rightarrow xdx=\frac{du}{2}.$

Now substituting these values we get,

$\;\;\;=\int\frac{dt/2}{t}-\int\frac{du/2}{u}.$

$\;\;\;=\frac{1}{2}\int\frac{dt}{t}-\frac{1}{2}\int\frac{du}{u}.$

On integrating we get,

$\;\;\;=\frac{1}{2}log|t|-\frac{1}{2}log|u|.$

$log a-log b=log|\frac{a}{b}|$,similarly

$\;\;\;=\frac{1}{2}log\frac{|t|}{|u|}+c.$

Substituting back for t and u we get,

$\int\frac{2x}{(x^2+1)(x^2+3)}=\frac{1}{2}log\frac{(x^2+1)}{(x^2+3)}+c.$

edited Jul 21, 2013