Evaluate the following integral:

Solution:

Let

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

Now,

Let $\frac{2 x+1}{x^{2}-1}=\frac{A}{x+1}+\frac{B}{x-1}$

$2 x+1=A(x-1)+B(x+1)$

Put $x=1$

$2+1=A \times 0+B \times 2$

$3=2 B$

$B=\frac{3}{2}$

Put $x=-1$

$-2+1=-2 A+B \times 0$

$-1=-2 \mathrm{~A}$

$A=\frac{1}{2}$

$I=\int x d x+\frac{1}{2} \int \frac{d x}{x+1}+\frac{3}{2} \int \frac{d x}{x-1}$

$\int \frac{\mathrm{dx}}{\mathrm{x}}=\log |\mathrm{x}|$ and $\int \mathrm{xdx}=\frac{\mathrm{x}^{2}}{2}$

Therefore,

$I=\frac{x^{2}}{2}+\frac{1}{2} \log |x+1|+\frac{3}{2} \log |x-1|+c$