Question

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

Answer 1

We have,

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

\[\text{Let }\frac{x^2 + 1}{\left( x - 2 \right)^2 \left( x + 3 \right)} = \frac{A}{x - 2} + \frac{B}{\left( x - 2 \right)^2} + \frac{C}{x + 3}\]

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

\[ \Rightarrow x^2 + 1 = A \left( x^2 - 2x + 3x - 6 \right) + B \left( x + 3 \right) + C \left( x^2 - 4x + 4 \right)\]

\[ \Rightarrow x^2 + 1 = A \left( x^2 + x - 6 \right) + B \left( x + 3 \right) + C \left( x^2 - 4x + 4 \right)\]

Equating coefficients of like terms

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

\[A + B - 4C = 0 ...................(2)\]

\[ - 6A + 3B + 4C = 1 .....................(3)\]

Solving (1), (2) and (3), we get

\[A = \frac{3}{5}, B = 1\text{ and }C = \frac{2}{5}\]

\[ \therefore I = \frac{3}{5}\int\frac{dx}{x - 2} + \int\frac{dx}{\left( x - 2 \right)^2} + \frac{2}{5}\int\frac{dx}{x + 3}\]

\[ = \frac{3}{5} \log \left| x - 2 \right| + \left[ \frac{\left( x - 2 \right)^{- 2 + 1}}{- 2 + 1} \right] + \frac{2}{5} \log \left| x + 3 \right| + C\]

\[ = \frac{3}{5}\log \left| x - 2 \right| - \frac{1}{x - 2} + \frac{2}{5} \log \left| x + 3 \right| + C\]