Question

Integrate the rational function: \(\frac {(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\)

Answer 1

\(\frac {(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\) = \(1- \frac {(4x^2+10)}{(x^2+3)(x^2+4)}\)

Let \(\frac {(4x^2+10)}{(x^2+3)(x^2+4)}\) = \(\frac {Ax+B}{(x^2+3)}+\frac {Cx+D}{(x^2+4)}\)

\(4x^2+10 = (Ax+B)(x^2+4)+(Cx+D)(x^2+3)\)

\(4x^2+10 = Ax^3+4Ax+Bx^2+4B+Cx^3+3Cx+Dx^2+3D\)

\(4x^2+10 = (A+C)x^3+(B+D)x^2+(4A+3C)x+(4B+3D)\)

\(Equating\  the\  coefficients\  of\  x^3, x^2, x,\  and\  constant \ term,\  we\  obtain\)
\(A + C = 0\)
\(B + D = 4\)
\(4A + 3C = 0\)
\(4B + 3D = 10\)
\(On\  solving\  these \ equations,\  we \ obtain\)
\(A = 0, B = −2, C = 0, \ and\  D = 6\)

∴ \(\frac {(4x^2+10)}{(x^2+3)(x^2+4)}\) =\(-\frac {2}{(x^2+3)}+\frac {6}{(x^2+4)}\)

\(\frac {(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\) = 1\(-\frac {2}{(x^2+3)}+\frac {6}{(x^2+4)}\)

⇒ \(∫\)\(\frac {(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\ dx\) = \(∫[{1+\frac {2}{(x^2+3)}-\frac {6}{(x^2+4)}}]dx\)

                                      = \(∫{1+\frac {2}{x^2+(\sqrt 3)^2}-\frac {6}{x^2+2^2}}dx\)

                                      = \(x+2(\frac {1}{\sqrt 3}tan^{-1}\frac {x}{\sqrt 3})-6(\frac 12 tan^{-1}\frac x2)+C\)

                                      =  \(x+\frac {2}{\sqrt 3}tan^{-1}\frac { x}{\sqrt 3}-3tan^{-1}\frac x2+C\)