Question

\[\int\frac{x^2}{\left( x - 1 \right) \left( x + 1 \right)^2} dx\]

Answer 1

We have,

\[I = \int\frac{x^2 dx}{\left( x - 1 \right) \left( x + 1 \right)^2}\]

\[\text{Let }\frac{x^2}{\left( x - 1 \right) \left( x + 1 \right)^2} = \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{\left( x + 1 \right)^2}\]

\[ \Rightarrow \frac{x^2}{\left( x - 1 \right) \left( x + 1 \right)^2} = \frac{A \left( x + 1 \right)^2 + B \left( x + 1 \right) \left( x - 1 \right) + C \left( x - 1 \right)}{\left( x + 1 \right)^2 \left( x - 1 \right)}\]

\[ \Rightarrow x^2 = A \left( x^2 + 2x + 1 \right) + B \left( x^2 - 1 \right) + C \left( x - 1 \right)\]

\[ \Rightarrow x^2 = \left( A + B \right) x^2 + x \left( 2A + C \right) + \left( A - B - C \right)\]

\[\text{Equating coefficients of like terms}\]

\[A + B = 1 ...................(1)\]

\[2A + C = 0 ....................(2)\]

\[A - B - C = 0 .......................(3)\]

\[\text{Solving (1), (2) and (3), we get}\]

\[A = \frac{1}{4}, B = \frac{3}{4}\text{ and }C = - \frac{1}{2}\]

\[ \therefore \frac{x^2}{\left( x - 1 \right) \left( x + 1 \right)^2} = \frac{1}{4 \left( x - 1 \right)} + \frac{3}{4 \left( x + 1 \right)} - \frac{1}{2 \left( x + 1 \right)^2}\]

\[ \Rightarrow I = \frac{1}{4}\int\frac{dx}{x - 1} + \frac{3}{4}\int\frac{dx}{x + 1} - \frac{1}{2}\int\frac{dx}{\left( x + 1 \right)^2}\]

\[ = \frac{1}{4} \log \left| x - 1 \right| + \frac{3}{4} \log \left| x + 1 \right| - \frac{1}{2} \times \frac{- 1}{x + 1} + C\]

\[ = \frac{1}{4}\log \left| x - 1 \right| + \frac{3}{4} \log \left| x + 1 \right| + \frac{1}{2 \left( x + 1 \right)} + C\]