Evaluate the following integral:

Here the denominator is already factored.

So let

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

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

$\Rightarrow 2 x+1=A(x-2)+B(x+1) \ldots \ldots($ 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 $x=2$ in the above equation, we get

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

$\Rightarrow 3 B=5$

$\Rightarrow \mathrm{B}=\frac{5}{3}$

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

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

$\Rightarrow-3 A=-1$

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

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

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

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

Split up the integral,

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

Let substitute $u=x+1 \Rightarrow d u=d x$ and $z=x-2 \Rightarrow d z=d x$, so the above equation becomes,

$\Rightarrow \frac{1}{3} \int\left[\frac{1}{\mathrm{u}}\right] \mathrm{du}+\frac{5}{3} \int\left[\frac{1}{\mathrm{z}}\right] \mathrm{dz}$

On integrating we get

$\Rightarrow \frac{1}{3} \log |\mathrm{u}|+\frac{5}{3} \log |\mathrm{z}|+\mathrm{C}$

Substituting back, we get

$\Rightarrow \frac{1}{3} \log |x+1|+\frac{5}{3} \log |x-2|+C$

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

Hence,

$\int \frac{2 x+1}{(x+1)(x-2)} d x=\frac{1}{3} \log |x+1|+\frac{5}{3} \log |x-2|+C$