Question

Evaluate the following integrals:

$\int \frac{x^3+x+1}{x^2-1}\mathrm{d}x$$\int \frac{x^3+x+1}{x^2-1}\mathrm{d}x$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x^3+x+1}{x^2-1}\mathrm{d}x \\ \mathrm{Here}\mathrm{the}\mathrm{integrand}\frac{x^3+x+1}{x^2-1}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{proper}\mathrm{rational}\mathrm{function},\mathrm{so}\mathrm{we}\mathrm{divide}{x}^3+x+1\mathrm{by}{x}^2-1\mathrm{and}\mathrm{find}\mathrm{that} \\ \frac{x^3+x+1}{x^2-1}=x+\frac{2x+1}{x^2-1}=x+\frac{2x+1}{\left(x+1\right)\left(x-1\right)} \\ \mathrm{Let}\frac{2x+1}{\left(x+1\right)\left(x-1\right)}=\frac{A}{x+1}+\frac{B}{x-1} \\ \Rightarrow 2x+1=A\left(x-1\right)+B\left(x+1\right) \\ \mathrm{Equating}\mathrm{the}\mathrm{coefficients}\mathrm{of}x\mathrm{and}\mathrm{constants},\mathrm{we}\mathrm{get} \\ 2=A+B\mathrm{and}1=-A+B \\ \mathrm{or}A=\frac{1}{2}\mathrm{and}B=\frac{3}{2} \\ \therefore I=\int \left(x+\frac{{\displaystyle \frac{1}{2}}}{x+1}+\frac{{\displaystyle \frac{3}{2}}}{x-1}\right)\mathrm{d}x \\ =\int x\mathrm{d}x+\frac{1}{2}\int \frac{1}{x+1}\mathrm{d}x+\frac{3}{2}\int \frac{1}{x-1}\mathrm{d}x \\ =\frac{x^2}{2}+\frac{1}{2}\log \left|x+1\right|+\frac{3}{2}\log \left|x-1\right|+c \\ =\frac{x^2}{2}+\frac{1}{2}\log \left|x+1\right|+\frac{3}{2}\log \left|x-1\right|+c \\ \mathrm{Hence},\int \frac{x^3+x+1}{x^2-1}\mathrm{d}x=\frac{x^2}{2}+\frac{1}{2}\log \left|x+1\right|+\frac{3}{2}\log \left|x-1\right|+c \end{gathered}$$