Evaluate the following integral:

Denominator is factorized, so let separate the fraction through partial fraction, hence let

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

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

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

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

We need to solve for A, B and C. 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

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

$\Rightarrow 1=6 A+0+0$

$\Rightarrow A=\frac{1}{6}$

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

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

$\Rightarrow 1=0-2 B+0$

$\Rightarrow \mathrm{B}=-\frac{1}{2}$

Now put $x=-2$ in equation (ii), we get

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

$\Rightarrow 1=0+0+3 C$

$\Rightarrow C=\frac{1}{3}$

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

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

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

$\Rightarrow \int\left[\frac{\frac{1}{6}}{(\mathrm{x}-1)}+\frac{-\frac{1}{2}}{\mathrm{x}+1}+\frac{\frac{1}{3}}{\mathrm{x}+2}\right] \mathrm{dx}$

Split up the integral,

$\Rightarrow \frac{1}{6} \int\left[\frac{1}{(x-1)}\right] d x-\frac{1}{2} \int\left[\frac{1}{x+1}\right] d x+\frac{1}{3} \int\left[\frac{1}{x+2}\right] d x$

Let substitute

$u=x-1 \Rightarrow d u=d x$,

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

$z=x+2 \Rightarrow d z=d x$, so the above equation becomes,

$\Rightarrow \frac{1}{6} \int\left[\frac{1}{\mathrm{u}}\right] \mathrm{du}-\frac{1}{2} \int\left[\frac{1}{\mathrm{y}}\right] \mathrm{dy}+\frac{1}{3} \int\left[\frac{1}{\mathrm{z}}\right] \mathrm{dz}$

On integrating we get

$\Rightarrow \frac{1}{6} \log |\mathrm{u}|-\frac{1}{2} \log |\mathrm{y}|+\frac{1}{3} \log |\mathrm{z}|+\mathrm{C}$

Substituting back, we get

$\Rightarrow \frac{1}{6} \log |x-1|-\frac{1}{2} \log |x+1|+\frac{1}{3} \log |x+2|+C$

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

Hence,

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

$=\frac{1}{6} \log |x-1|-\frac{1}{2} \log |x+1|+\frac{1}{3} \log |x+2|+C$