Evaluate the following integral:

Solution:

First we simplify numerator, we get

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

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

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

$=1+\frac{2}{(x-1)(x+1)}$

Now the denominator is factorized, so let separate the fraction through partial fraction, hence let

$\frac{2}{(x+1)(x-1)}=\frac{A}{x+1}+\frac{B}{x-1} \ldots \ldots$ (i)

$\Rightarrow \frac{2}{(x+2)(x-1)}=\frac{A(x-1)+B(x+1)}{(x+2)(x-1)}$

$\Rightarrow 2=A(x-1)+B(x+1) \ldots \ldots$ (ii)

We need to solve for $A$ and $B$. One way to do this is to pick values for $x$ which will cancel each variable.

Put $x=1$ in the above equation, we get

$\Rightarrow 2=A(1-1)+B(1+1)$

$\Rightarrow 2=0+2 B$

$\Rightarrow B=1$

Now put $x=-1$ in equation (ii), we get

$\Rightarrow 2=A((-1)-1)+B((-1)+1)$

$\Rightarrow 2=-2 A+0$

$\Rightarrow A=-1$

We put the values of $A$ and $B$ values back into our partial fractions in equation (i) and replace this as the integrand. We get

$\int\left[1+\frac{2}{(x-1)(x+1)}\right] d x$

$\Rightarrow \int\left[1+\frac{A}{x+1}+\frac{B}{x-1}\right] d x$

$\Rightarrow \int\left[1+\frac{-1}{x+1}+\frac{1}{x-1}\right] d x$

Split up the integral,

$\Rightarrow \int 1 \mathrm{dx}-\int\left[\frac{1}{\mathrm{x}+1}\right] \mathrm{dx}+\int\left[\frac{1}{\mathrm{x}-1}\right] \mathrm{dx}$

Let substitute $u=x+1 \Rightarrow d u=d x$ and $z=x-1 \Rightarrow d z=d x$, so the above equation becomes,

$\Rightarrow \int 1 \mathrm{dx}-\int\left[\frac{1}{\mathrm{u}}\right] \mathrm{du}+\int\left[\frac{1}{\mathrm{z}}\right] \mathrm{dz}$

On integrating we get

$\Rightarrow x-\log |u|+\log |z|+C$

Substituting back, we get

$\Rightarrow x-\log |x+1|+\log |x-1|+C$

Applying the logarithm rule we get

$\Rightarrow x+\log \left|\frac{x-1}{x+1}\right|+C$

Note: the absolute value signs account for the domain of the natural log function $(x>0)$.

Hence,

$\int \frac{x^{2}+1}{x^{2}-1} d x=x+\log \left|\frac{x-1}{x+1}\right|+C$