Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{\left(x^2+1\right)dx}{{\left(x-2\right)}^2\left(x+3\right)} \\ \mathrm{Let}\frac{x^2+1}{{\left(x-2\right)}^2\left(x+3\right)}=\frac{A}{x-2}+\frac{B}{{\left(x-2\right)}^2}+\frac{C}{x+3} \\ \Rightarrow \frac{x^2+1}{{\left(x-2\right)}^2\left(x+3\right)}=\frac{A\left(x-2\right)\left(x+3\right)+B\left(x+3\right)+C{\left(x-2\right)}^2}{{\left(x-2\right)}^2\left(x+3\right)} \\ \Rightarrow x^2+1=A\left(x^2-2x+3x-6\right)+B\left(x+3\right)+C\left(x^2-4x+4\right) \\ \Rightarrow x^2+1=A\left(x^2+x-6\right)+B\left(x+3\right)+C\left(x^2-4x+4\right) \\ \mathrm{Equating}\mathrm{coefficients}\mathrm{of}\mathrm{like}\mathrm{terms} \\ A+C=1.....\left(1\right) \\ A+B-4C=0.....\left(2\right) \\ -6A+3B+4C=1.....\left(3\right) \\ \mathrm{Solving}\left(1\right),\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ A=\frac{3}{5},B=1\mathrm{and}C=\frac{2}{5} \\ \therefore I=\frac{3}{5}\int \frac{dx}{x-2}+\int \frac{dx}{{\left(x-2\right)}^2}+\frac{2}{5}\int \frac{dx}{x+3} \\ =\frac{3}{5}\log \left|x-2\right|+\left[\frac{{\left(x-2\right)}^{-2+1}}{-2+1}\right]+\frac{2}{5}\log \left|x+3\right|+C \\ =\frac{3}{5}\log \left|x-2\right|-\frac{1}{x-2}+\frac{2}{5}\log \left|x+3\right|+C \end{gathered}$$