Evaluate the following integral:
$\int \frac{1}{x \log x(2+\log x)} d x$
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,
$\int \frac{1}{x \log x(2+\log x)} d x=\frac{1}{2} \log \left|\frac{\log x}{2+\log x}\right|+C$
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