Evaluate the following integral:

Here the denominator is already factored.

So let

$\frac{1}{x(x-2)(x-4)}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x-4} \ldots \ldots(i)$

$\Rightarrow \frac{1}{x(x-2)(x-4)}=\frac{A(x-2)(x-4)+B x(x-4)+C x(x-2)}{x(x-2)(x-4)}$

$\Rightarrow 1=A(x-2)(x-4)+B x(x-4)+C x(x-2) \ldots \ldots$ (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

$\Rightarrow 1=A(0-2)(0-4)+B(0)(0-4)+C(0)(0-2)$

$\Rightarrow 1=8 A+0+0$

$\Rightarrow \mathrm{A}=\frac{1}{8}$

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

$\Rightarrow 1=A(2-2)(2-4)+B(2)(2-4)+C(2)(2-2)$

$\Rightarrow 1=0-4 B+0$

$\Rightarrow B=-\frac{1}{4}$'

Now put $x=4$ in equation (ii), we get

$\Rightarrow 1=A(4-2)(4-4)+B(4)(4-4)+C(4)(4-2)$

$\Rightarrow 1=0+0+8 C$

$\Rightarrow C=\frac{1}{8}$

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

$\int\left[\frac{A}{x}+\frac{B}{x-2}+\frac{C}{x-4}\right] d x$

$\Rightarrow \int\left[\frac{\frac{1}{8}}{x}+\frac{-\frac{1}{4}}{x-2}+\frac{\frac{1}{8}}{x-4}\right] d x$

Split up the integral,

$\Rightarrow \frac{1}{8} \int\left[\frac{1}{x}\right] d x-\frac{1}{4} \int\left[\frac{1}{x-2}\right] d x+\frac{1}{8} \int\left[\frac{1}{x-4}\right] d x$

Let substitute $u=x-4 \Rightarrow d u=d x$ and $z=x-2 \Rightarrow d z=d x$, so the above equation becomes,

$\Rightarrow \frac{1}{8} \int\left[\frac{1}{x}\right] d x-\frac{1}{4} \int\left[\frac{1}{z}\right] d z+\frac{1}{8} \int\left[\frac{1}{u}\right] d u$

On integrating we get

$\Rightarrow \frac{1}{8} \log |x|-\frac{1}{4} \log |z|+\frac{1}{8} \log |u|+C$

Substituting back, we get

$\Rightarrow \frac{1}{8} \log |x|-\frac{1}{4} \log |x-2|+\frac{1}{8} \log |x-4|+C$

We will take $\frac{1}{8}$ common, we get

$\Rightarrow \frac{1}{8}[\log |x|-2 \log |x-2|+\log |x-4|+C]$

Applying the logarithm rule we can rewrite the above equation as

$\Rightarrow \frac{1}{8}\left[\log \left|\frac{x}{(x-2)^{2}}\right|+\log |x-4|+C\right]$

$\Rightarrow \frac{1}{8}\left[\log \left|\frac{x(x-4)}{(x-2)^{2}}\right|\right]+C$

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

Hence,

$\int \frac{1}{x(x-2)(x-4)} d x=\frac{1}{8}\left[\log \left|\frac{x(x-4)}{(x-2)^{2}}\right|\right]+C$