# Evaluate the following integral:

Question:

Evaluate the following integral:

$\int \frac{x^{2}+6 x-8}{x^{3}-4 x} d x$

Solution:

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

$\frac{x^{2}+6 x-8}{x^{3}-4 x}$

$=\frac{x^{2}+6 x-8}{x\left(x^{2}-4\right)}$

$\frac{\mathrm{x}^{2}+6 \mathrm{x}-8}{\mathrm{x}(\mathrm{x}-2)(\mathrm{x}+2)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{B}}{\mathrm{x}-2}+\frac{\mathrm{C}}{\mathrm{x}+2} \ldots \ldots$ (i)

$\Rightarrow \frac{x^{2}+6 x-8}{x(x-2)(x+2)}=\frac{A(x-2)(x+2)+B x(x+2)+C x(x-2)}{x(x-2)(x+2)}$

$\Rightarrow x^{2}+6 x-8=A(x-2)(x+2)+B x(x+2)+C x(x-2) \ldots \ldots$ (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 $x=0$ in the above equation, we get

$\Rightarrow 0^{2}+6(0)-8=A(0-2)(0+2)+B(0)(0+2)+C(0)(0-2)$

$\Rightarrow-8=-4 A+0+0$

$\Rightarrow A=2$

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

$\Rightarrow 2^{2}+6(2)-8=A(2-2)(2+2)+B(2)(2+2)+C(2)(2-2)$

$\Rightarrow 8=0+8 B+0$

$\Rightarrow B=1$

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

$\Rightarrow(-2)^{2}+6(-2)-8=A((-2)-2)((-2)+2)+B(-2)((-2)+2)+C(-2)((-2)-2)$

$\Rightarrow-16=0+0+8 C$

$\Rightarrow C=-2$

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

$\int\left[\frac{\mathrm{x}^{2}+6 \mathrm{x}-8}{\mathrm{x}(\mathrm{x}-2)(\mathrm{x}+2)}\right] \mathrm{dx}$

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

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

Split up the integral,

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

Let substitute

$u=x-2 \Rightarrow d u=d x$

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

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

On integrating we get

$\Rightarrow 2 \log |x|+\log |u|-2 \log |y|+C$

Substituting back, we get

$\Rightarrow \log |x|+\log |x-2|-2 \log |x+2|+C$

Applying logarithm rule, we get

$\Rightarrow \log |x(x-2)|-\log \left|(x+2)^{2}\right|+C$

$\Rightarrow \log \left|\frac{x(x-2)}{(x+2)^{2}}\right|+C$

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

Hence,

$\int \frac{x^{2}+6 x-8}{x(x-2)(x+2)} d x=\log \left|\frac{x(x-2)}{(x+2)^{2}}\right|+C$