Question

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

Answer 1

$$\begin{gathered} \int \left(\frac{x^2+1}{x^2-1}\right)dx \\ =\int \left(\frac{x^2-1+2}{x^2-1}\right)dx \\ =\int dx+2\int \frac{1}{x^2-1^2}dx \\ =\int dx+2\int \frac{1}{\left(x-1\right)\left(x+1\right)}dx...\left(1\right) \\ \therefore \frac{1}{\left(x-1\right)\left(x+1\right)}=\frac{A}{x-1}+\frac{B}{x+1} \\ \Rightarrow \frac{1}{\left(x-1\right)\left(x+1\right)}=\frac{A\left(x+1\right)+B\left(x-1\right)}{\left(x-1\right)\left(x+1\right)} \\ \Rightarrow 1=A\left(x+1\right)B\left(x-1\right)....\left(2\right) \\ \mathrm{Putting}x+1=0\mathrm{or}x=-1\mathrm{in}\mathrm{eq}\left(2\right) \\ \Rightarrow 1=A\times 0+B\left(-1-1\right) \\ \Rightarrow B=\frac{-1}{2} \\ \mathrm{Putting}x-1=0\mathrm{or}x=1\mathrm{in}\mathrm{eq}\left(2\right) \\ \Rightarrow 1=A\left(1+1\right)+B\times 0 \\ \Rightarrow A=\frac{1}{2} \\ \therefore \frac{1}{\left(x-1\right)\left(x+1\right)}=\frac{1}{2\left(x-1\right)}-\frac{1}{2\left(x+1\right)}...\left(3\right) \\ \mathrm{From}\mathrm{eq}\left(1\right)\&\mathrm{eq}\left(3\right) \\ \int \left(\frac{x^2+1}{x^2-1}\right)dx=\int dx+2\int \left[\frac{1}{2\left(x-1\right)}-\frac{1}{2\left(x+1\right)}\right]dx \\ =\int dx+\int \frac{dx}{x-1}-\int \frac{dx}{x+1} \\ =x+\ln \left|x-1\right|=-\ln \left|x+1\right|+C \\ =x+\ln \left|\frac{x-1}{x+1}\right|+C \end{gathered}$$