Question

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

Answer 1

∴$\frac {1}{(x^2+1)(x^2+4)}$ = $\frac {Ax+B}{(x^2+1)}+\frac {Cx+D}{(x^2+4)}$

$⇒1 = (Ax+B)(x^2+4)+(Cx+D)(x^2+1)$

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

Equating the coefficients of $x^3,x^2,x,$ and constant term,we obtain

$A+C=0$

$B+D=0$

$4A+C=0$

$4B+D=1$

On solving these equations, we obtain

$A=0,\ B=\frac 13,\ C=0,\ D=-\frac 13$

From equation(1), we obtain

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

∫$$\frac {1}{(x^2+1)(x^2+4)} = $\frac 13∫\frac {1}{x^2+1}dx-\frac {1}3∫\frac {1}{x^2+4}dx$

                              =$\frac 13\tan^{-1}x-\frac 13.\frac 12tan^{-1}\frac x2+C$

                              =$\frac 13tan^{-1}x-\frac 16tan^{-1}\frac x2+C$