Question

$\int \frac{x^2+1}{x^2-5x+6}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I\int \left(\frac{x^2+1}{x^2-5x+6}\right)dx \\ \mathrm{Dividing}\mathrm{Numerator}\mathrm{by}\mathrm{Denominator} \\ {x}^2-5x+6\stackrel{1}{\overline{)x^2+1}} \\ {x}^2-5x+6 \\ \underline{-+-} \\ 5x-5 \\ \frac{x^2+1}{x^2-5x+6}=1+\left(\frac{5x-5}{x^2-5x+6}\right).....\left(1\right) \\ \mathrm{Also}\frac{5x-5}{x^2-5x+6}=\frac{5x-5}{\left(x-2\right)\left(x-3\right)} \\ \mathrm{Let}\frac{5x-5}{\left(x-2\right)\left(x-3\right)}=\frac{A}{x-2}+\frac{B}{x-3} \\ \Rightarrow \frac{5x-5}{\left(x-2\right)\left(x-3\right)}=\frac{A\left(x-3\right)+B\left(x-2\right)}{\left(x-2\right)\left(x-3\right)} \\ \Rightarrow 5x-5=A\left(x-3\right)+B\left(x-2\right) \\ \mathrm{let}x=3 \\ 5\times 3-5=A\times 0+B\left(3-2\right) \\ 10=B \\ \mathrm{let}x=2 \\ 5\times 2-5=A\left(2-3\right)+B\times 0 \\ A=-5 \\ \frac{5x-5}{\left(x-2\right)\left(x-3\right)}=\frac{-5}{x-2}+\frac{10}{x-3}.....\left(2\right) \\ \mathrm{from}\left(1\right)\mathrm{and}\left(2\right) \\ I=\int dx-5\int \frac{dx}{x-2}+10\int \frac{dx}{x-3} \\ =x-5\log \left|x-2\right|+10\log \left|x-3\right|+C \end{gathered}$$