Question

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

Answer 1

We have,

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

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

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

\[ \Rightarrow x^2 + x - 1 = A \left( x^2 + 3x + 2 \right) + B \left( x + 2 \right) + C \left( x^2 + 2x + 1 \right)\]

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

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

\[3A + B + 2C = 1 ...................(2)\]

\[2A + 2B + C = - 1 .......................(3)\]

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

\[A = 0 \]

\[B = - 1\]

\[C = 1\]

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

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

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

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

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