# Evaluate the following integral:

Question:

Evaluate the following integral:

$\int \frac{5 x}{(x+1)\left(x^{2}-4\right)} d x$

Solution:

$\frac{5 x}{(x+1)\left(x^{2}-4\right)}=\frac{5 x}{(x+1)(x-2)(x+2)}$

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

$\frac{5 x}{(x+1)(x-2)(x+2)}=\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{x+2} \ldots \ldots$ (i)

$\Rightarrow \frac{5 x}{(x+1)(x-2)(x+2)}$

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

$\Rightarrow 5 x=A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2) \ldots \ldots($ 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

$\Rightarrow 5(-1)=A((-1)-2)((-1)+2)+B((-1)+1)((-1)+2)+C((-1)+1)((-1)-2)$

$\Rightarrow-5=-3 \mathrm{~A}+0+0$

$\Rightarrow A=\frac{5}{3}$

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

$\Rightarrow 5(-2)=A((-2)-2)((-2)+2)+B((-2)+1)((-2)+2)+C((-2)+1)((-2)-2)$

$\Rightarrow-10=0+0+4 C$

$\Rightarrow C=-\frac{10}{4}=-\frac{5}{2}$

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

$\Rightarrow 5(2)=A((2)-2)((2)+2)+B((2)+1)((2)+2)+C((2)+1)((2)-2)$

$\Rightarrow 10=0+12 B+0$

$\Rightarrow \mathrm{B}=\frac{10}{12}=\frac{5}{6}$

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

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

$\Rightarrow \int\left[\frac{\frac{5}{3}}{x+1}+\frac{-\frac{5}{2}}{x-2}+\frac{\frac{5}{6}}{x+2}\right] d x$

Split up the integral,

$\Rightarrow \frac{5}{3} \int\left[\frac{1}{x+1}\right] d x-\frac{5}{2} \int\left[\frac{1}{x-2}\right] d x+\frac{5}{6} \int\left[\frac{1}{x+2}\right] d x$

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

$\Rightarrow \frac{5}{3} \int\left[\frac{1}{\mathrm{u}}\right] \mathrm{du}-\frac{5}{2} \int\left[\frac{1}{\mathrm{y}}\right] \mathrm{dy}+\frac{5}{6} \int\left[\frac{1}{\mathrm{z}}\right] \mathrm{dz}$

On integrating we get

$\Rightarrow \frac{5}{3} \log |\mathrm{u}|-\frac{5}{2} \log |\mathrm{y}|+\frac{5}{6} \log |\mathrm{z}|+\mathrm{C}$

Substituting back, we get

$\Rightarrow \frac{5}{3} \log |x+1|-\frac{5}{2} \log |x-2|+\frac{5}{6} \log |x+2|+C$

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

Hence,

$\int \frac{5 x}{(x+1)\left(x^{2}-4\right)} d x=\frac{5}{3} \log |x+1|-\frac{5}{2} \log |x-2|+\frac{5}{6} \log |x+2|+C$