Question

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

Answer 1

We have,

\[I = \int\frac{\left( x^2 + x + 1 \right)}{\left( x + 1 \right)^2 \left( x + 2 \right)}dx\]
\[\text{Let }\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} = \frac{A}{x + 1} + \frac{B}{\left( x + 1 \right)^2} + \frac{C}{x + 2}\]
\[ \Rightarrow \frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} = \frac{A\left( x + 1 \right) \left( x + 2 \right) + B\left( x + 2 \right) + C \left( x + 1 \right)^2}{\left( x + 1 \right)^2 \left( x + 2 \right)}\]
\[ \Rightarrow x^2 + x + 1 = A\left( x^2 + x + 2x + 2 \right) + Bx + 2B + C\left( x^2 + 2x + 1 \right)\]
\[ \Rightarrow x^2 + x + 1 = \left( A + C \right) x^2 + \left( 3A + B + 2C \right)x + \left( 2A + 2B + C \right)\]
\[\text{Equating coefficient of like terms} . \]
\[A + C = 1 . . . . . \left( 1 \right)\]
\[3A + B + 2C = 1 . . . . . \left( 2 \right)\]
\[2A + 2B + C = 1 . . . . . \left( 3 \right)\]
\[\text{Solving these three equations we get}\]
\[A = - 2\]
\[B = 1\]
\[C = 3\]
\[\text{Hence, }\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} = \frac{- 2}{x + 1} + \frac{1}{\left( x + 1 \right)^2} + \frac{3}{x + 2}\]
\[ \therefore I = - 2\int\frac{dx}{x + 1} + \int\frac{d}{\left( x + 1 \right)^2} + 3\int\frac{dx}{x + 2}\]
\[ = - 2 \log \left| x + 1 \right| - \frac{1}{x + 1} + 3 \log \left| x + 2 \right| + C\]