# Integrate the function$\int\frac{1}{(x^2+1)(x^2+4)}$

Toolbox:
• (i)An rational function is of the form. $\frac{1}{(x^2+a)(x^2+b)}$ can be written as $\frac{Ax+B}{(x^2+a)}+\frac{Cx+D}{(x^2+b)}$
• (ii)$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan ^{-1}(x/a)+c$
Given $I=\int\frac{1}{(x^2+1)(x^2+4)}dx$

Consider $\frac{1}{(x^2+1)(x^2+4)},$ This can be resolved as

$\large\frac{(Ax+B)}{(X^2+1)}+\frac{(x+D)}{x^2+4}$

$1=(Ax+B)(x^2+4)+(Cx+D)(x^2+1)$

$1=Ax^3+4Ax+Bx^2+4B+Cx^3+Cx+Dx^2+D$

Equating the coefficient of $x^3$

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

Equating the coefficient of $x^2$

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

Equating the coefficient of $x$

$0=4A+C$-----(3)

Equating the constant term,

$1=4B+D$-----(4)

Subtract equ(2) and equ(4)

$B+D=0$
$4B+D=1$
___________
$-3B=-1\qquad B=\frac{1}{3}$

Substituting for B in equation (2)

$B+D=0\qquad Therefore\;\frac{1}{3}+D=0$

$=>D=-\frac{1}{3}$

Consider equ (1) and equ(3)

$A+C=0$

$4A+C=0$

We can understand that A and C=0

Hence $A=0,B=\frac{1}{3},c=0\; and \; d=-\frac{1}{3}$

Hence $\frac{1}{(x^2+1)(x^2+4)}=\frac{1}{3(x^2+1)}-\frac{1}{3(x^2+4)}$

Therefore $\frac{dx}{(x^2+1)(x^2+4)}=\frac{1}{3}\int \frac{dx}{x^2+1}-\frac{1}{3}\int \frac{1}{x^2+4}dx$

This is of the form $\int \frac{dx}{a^2+x^2}=\frac{1}{a} \tan ^{-1}(\frac{x}{a})+c$

On integrating we get,

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

$I=\frac{1}{3}\tan ^{-1}(x)-\frac{1}{6} \tan ^{-1}\frac{x}{2}+c$