Evaluate the following integral:

Let substitute $u=\log x \Rightarrow d u=\frac{1}{x} d x$, so the given equation becomes

$\int \frac{1}{x \log x(2+\log x)} d x=\int \frac{1}{u(2+u)} d u \ldots .(i)$

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

$\frac{1}{u(2+u)}=\frac{A}{u}+\frac{B}{(2+u)} \ldots$(ii)

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

$\Rightarrow 1=\mathrm{A}(2+\mathrm{u})+\mathrm{Bu} \ldots \ldots(\mathrm{ii})$

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 $u=-2$ in above equation, we get

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

$\Rightarrow 1=-2 B$

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

Now put $u=0$ in equation (ii), we get

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

$\Rightarrow 1=2 A+0$

$\Rightarrow A=\frac{1}{2}$

We put the values of $A$ and $B$ values back into our partial fractions in equation (ii) and replace this as the integrand. We get

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

$\Rightarrow \int\left[\frac{\mathrm{A}}{\mathrm{u}}+\frac{\mathrm{B}}{(2+\mathrm{u})}\right] \mathrm{du}$

$\Rightarrow \int\left[\frac{\frac{1}{2}}{\mathrm{u}}+\frac{-\frac{1}{2}}{(2+\mathrm{u})}\right] \mathrm{du}$

Split up the integral,

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

Let substitute

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

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

On integrating we get

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

Substituting back the value of $z$, we get

$\Rightarrow \frac{1}{2} \log |\mathrm{u}|-\frac{1}{2} \log |2+\mathrm{u}|+\mathrm{C}$

Now substitute back the value of $u$, we get

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

Applying the rules of logarithm we get

$\Rightarrow \frac{1}{2} \log \left|\frac{\log x}{2+\log x}\right|+C$

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

Hence,

$\int \frac{1}{x \log x(2+\log x)} d x=\frac{1}{2} \log \left|\frac{\log x}{2+\log x}\right|+C$