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

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

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

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

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