Evaluate the following integral:

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

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

$\frac{2 x-3}{(x-1)(x+1)(2 x+3)}=\frac{A}{(x-1)}+\frac{B}{x+1}+\frac{C}{2 x+3} \ldots \ldots$ (i)

$\Rightarrow \frac{2 x-3}{(x-1)(x+1)(2 x+3)}$

$=\frac{A(x+1)(2 x+3)+B(x-1)(2 x+3)+C(x-1)(x+1)}{(x-1)(x+1)(2 x+3)}$

$\Rightarrow 2 x-3=A(x+1)(2 x+3)+B(x-1)(2 x+3)+C(x-1)(x+1) \ldots \ldots$ (ii)

We need to solve for $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$. One way to do this is to pick values for $\mathrm{x}$ which will cancel each variable.

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

$\Rightarrow 2(-1)-3=A((-1)+1)(2(-1)+3)+B((-1)-1)(2(-1)+3)+C((-1)-1)((-1)+1)$

$\Rightarrow-5=0-2 B+0$

$\Rightarrow B=\frac{5}{2}$

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

$\Rightarrow 2(1)-3=A((1)+1)(2(1)+3)+B((1)-1)(2(1)+3)+C((1)-1)((1)+1)$

$\Rightarrow-1=10 A+0+0$

$\Rightarrow A=-\frac{1}{10}$

Now put $\mathrm{x}=-\frac{3}{2}$ in equation (ii), we get

$\Rightarrow 2\left(-\frac{3}{2}\right)-3$

$=A\left(\left(-\frac{3}{2}\right)+1\right)\left(2\left(-\frac{3}{2}\right)+3\right)$

$+B\left(\left(-\frac{3}{2}\right)-1\right)\left(2\left(-\frac{3}{2}\right)+3\right)+C\left(\left(-\frac{3}{2}\right)-1\right)\left(\left(-\frac{3}{2}\right)+1\right)$

$\Rightarrow-6=0+0+\frac{5}{4} C$

$\Rightarrow C=-\frac{24}{5}$

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

$\int\left[\frac{2 x-3}{(x-1)(x+1)(2 x+3)}\right] d x$

$\Rightarrow \int\left[\frac{A}{(x-1)}+\frac{B}{x+1}+\frac{C}{2 x+3}\right] d x$

$\Rightarrow \int\left[\frac{-\frac{1}{10}}{(x-1)}+\frac{\frac{5}{2}}{x+1}+\frac{-\frac{24}{5}}{2 x+3}\right] d x$

Split up the integral,

$\Rightarrow-\frac{1}{10} \int\left[\frac{1}{x-1}\right] d x+\frac{5}{2} \int\left[\frac{1}{x+1}\right] d x-\frac{24}{5} \int\left[\frac{1}{2 x+3}\right] d x$

Let substitute

$u=x+1 \Rightarrow d u=d x$

$y=x-1 \Rightarrow d y=d x$ and

$\mathrm{z}=2 \mathrm{x}+3 \Rightarrow \mathrm{dz}=2 \mathrm{dx} \Rightarrow \mathrm{dx}=\frac{\mathrm{dz}}{2}$ so the above equation becomes,

$\Rightarrow-\frac{1}{10} \int\left[\frac{1}{y}\right] \mathrm{dy}+\frac{5}{2} \int\left[\frac{1}{\mathrm{u}}\right] \mathrm{du}-\frac{24}{5} \int \frac{\left[\frac{1}{\mathrm{z}}\right] \mathrm{dz}}{2}$

On integrating we get

$\Rightarrow-\frac{1}{10} \log |y|+\frac{5}{2} \log |u|-\frac{12}{5} \log |z|+C$

Substituting back, we get

$\Rightarrow-\frac{1}{10} \log |x-1|+\frac{5}{2} \log |x+1|-\frac{12}{5} \log |2 x+3|+c$

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

Hence,

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

$=-\frac{1}{10} \log |x-1|+\frac{5}{2} \log |x+1|-\frac{12}{5} \log |2 x+3|+C$