Question

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

Answer 1

\[\text{We have}, \]
\[I = \int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} \text{ dx}\]
\[\text{ Let }\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} = \frac{A}{x + 1} + \frac{B}{\left( x + 1 \right)^2} + \frac{C}{x + 2} . . . . . \left( 1 \right)\]
\[ \Rightarrow x^2 + x + 1 = A\left( x + 1 \right)\left( x + 2 \right) + B\left( x + 2 \right) + C \left( x + 1 \right)^2 . . . . . \left( 2 \right)\]
\[ \text{ Putting x = - 1 in }\left( 2 \right), \text{we get}\]
\[ B = 1\]
\[ \text{ Putting x = - 2 in }\left( 2 \right), \text{we get}\]
\[ C = 3\]
\[ \text{ Putting x = 0 in} \left( 2 \right), \text{we get}\]
\[1 = 2A + 2B + C\]
\[ \Rightarrow 1 = 2A + 2 + 3\]
\[ \Rightarrow - 4 = 2A\]
\[ \Rightarrow A = - 2\]
\[\text{Now}, \left( 1 \right) \text{becomes}\]
\[\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} = \frac{- 2}{x + 1} + \frac{1}{\left( x + 1 \right)^2} + \frac{3}{x + 2}\]
\[\text{Therefore, integral becomes}\]
\[I = \int\left[ \frac{- 2}{x + 1} + \frac{1}{\left( x + 1 \right)^2} + \frac{3}{x + 2} \right]dx\]
\[ = - 2 \text{ log} \left| x + 1 \right| - \frac{1}{\left( x + 1 \right)} + 3 \text{ log} \left| x + 2 \right| + C\]