Question

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

Answer 1

Let \(\frac {2}{(1-x)(1+x^2)}\) \(=\) \(\frac {A}{(1-x)}+\frac {Bx+C}{(1+x^2)}\)

\(2 = A(1+x^2)+(Bx+C)(1-x)\)

\(2 = A+Ax^2+Bx-Bx+C-Cx\)

Equating the coefficient of x², x, and constant term, we obtain
\(A − B = 0\)
\(B − C = 0\)
\(A + C = 2\)
On solving these equations, we obtain
\(A = 1, \ B = 1, \ and \ C = 1\)

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

⇒ \(∫\)\(\frac {2}{(1-x)(1+x^2)}\) = \(∫\)\(\frac {1}{1-x}\ dx\)+ \(∫\)\(\frac {x}{1+x^2}\ dx\) + \(∫\)\(\frac {1}{1+x^2}\ dx\)

                                      = - \(∫\)\(\frac {1}{1-x}\ dx\) + \(\frac 12\)\(∫\)\(\frac {2x}{1+x^2}\ dx\) + \(∫\)\(\frac {1}{1+x^2}\ dx\)

                                      = -\(log\ |x-1|+\frac 12log|1+x^2|+tan^{-1}x+C\)