Question

\[\int\frac{x^2}{\left( x - 1 \right)^3 \left( x + 1 \right)} \text{ dx}\]

Answer 1

\[\text{We have}, \]
\[I = \int\frac{x^2}{\left( x - 1 \right)^3 \left( x + 1 \right)} \text{ dx}\]
\[\text{ Let } \frac{x^2}{\left( x - 1 \right)^3 \left( x + 1 \right)} = \frac{A}{x - 1} + \frac{B}{\left( x - 1 \right)^2} + \frac{C}{\left( x - 1 \right)^3} + \frac{D}{x + 1} . . . . . \left( 1 \right)\]
\[ \Rightarrow x^2 = A \left( x - 1 \right)^2 \left( x + 1 \right) + B\left( x - 1 \right)\left( x + 1 \right) + C \left( x + 1 \right) + D \left( x - 1 \right)^3 . . . . . \left( 2 \right)\]
\[\text{ Putting x }= 1 \text{ in }\left( 2 \right), \text{we get}\]
\[1 = 2C\]
\[ \Rightarrow C = \frac{1}{2}\]
\[\text{ Putting x = - 1 in} \left( 2 \right), \text{we get}\]
\[1 = - 8D\]
\[ \Rightarrow D = \frac{- 1}{8}\]

\[\text{ Putting x = 2 in}\left( 2 \right), \text{ we get}\]

\[4 = 3A + 3B + 3C + D\]

\[ \Rightarrow 4 = 3A + 3B + \frac{3}{2} - \frac{1}{8}\]

\[ \Rightarrow 3A + 3B = 4 - \frac{3}{2} + \frac{1}{8}\]

\[ \Rightarrow 3A + 3B = \frac{32 - 12 + 1}{8}\]

\[ \Rightarrow 3A + 3B = \frac{21}{8}\]

\[ \Rightarrow A + B = \frac{7}{8}\]

\[\text{ And putting x = 0 in} \left( 2 \right), \text{ we get}\]

\[0 = A - B + C - D\]

\[ \Rightarrow 0 = A - B + \frac{1}{2} + \frac{1}{8} ..................\left[ \because C = \frac{1}{2}, D = \frac{1}{8} \right]\]

\[ \Rightarrow A - B = - \frac{5}{8}\]

\[\]

 

 

\[\text{ Here, A + B} = \frac{7}{8} \text{ and A - B} = - \frac{5}{8} \Rightarrow A = \frac{1}{8} \text{ and B } = \frac{3}{4}\]

\[\text{ Therefore,} \left( 1 \right) \text{ becomes,} \]

\[\frac{x^2}{\left( x - 1 \right)^3 \left( x + 1 \right)} = \frac{1}{8\left( x - 1 \right)} + \frac{3}{4 \left( x - 1 \right)^2} + \frac{1}{2 \left( x - 1 \right)^3} - \frac{1}{8\left( x + 1 \right)}\]

\[\text{ Now, integral becomes}\]

\[I = \int\left[ \frac{1}{8\left( x - 1 \right)} + \frac{3}{4 \left( x - 1 \right)^2} + \frac{1}{2 \left( x - 1 \right)^3} - \frac{1}{8\left( x + 1 \right)} \right]dx\]

\[ = \frac{1}{8}\text{ log }\left| x - 1 \right| - \frac{3}{4\left( x - 1 \right)} - \frac{1}{4 \left( x - 1 \right)^2} - \frac{1}{8}\text{ log }\left| x + 1 \right| + C\]

\[ = \frac{1}{8}\text{ log }\left| \frac{x - 1}{x + 1} \right| - \frac{3}{4\left( x - 1 \right)} - \frac{1}{4 \left( x - 1 \right)^2} + C\]