Evaluate the following integral:
$\int \frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)} d x$
$\frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)}=\frac{2 x-3}{(x-1)(x+1)(2 x+3)}$
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,
$\int \frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)} d x$
$=-\frac{1}{10} \log |x-1|+\frac{5}{2} \log |x+1|-\frac{12}{5} \log |2 x+3|+C$
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