Question

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

Answer 1

We have,

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

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

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

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

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

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

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

\[A + C = 0 .................(1)\]

\[A + B - 2C = 1 ..................(2)\]

\[ - 2A + 2B + C = 0 .....................(3)\]

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

\[A = \frac{2}{9}, B = \frac{1}{3}\text{ and }C = - \frac{2}{9}\]

\[ \therefore \frac{x}{\left( x - 1 \right)^2 \left( x + 2 \right)} = \frac{2}{9 \left( x - 1 \right)} + \frac{1}{3 \left( x - 1 \right)^2} - \frac{2}{9 \left( x + 2 \right)}\]

\[ \Rightarrow I = \frac{2}{9}\int\frac{dx}{x - 1} + \frac{1}{3}\int\frac{dx}{\left( x - 1 \right)^2} - \frac{2}{9}\int\frac{dx}{x + 2}\]

\[ = \frac{2}{9} \log \left| x - 1 \right| + \frac{1}{3} \times \left( \frac{- 1}{x - 1} \right) - \frac{2}{9} \log \left| x + 2 \right| + C\]

\[ = \frac{2}{9}\log \left| \frac{x - 1}{x + 2} \right| - \frac{1}{3 \left( x - 1 \right)} + C\]