Question

If $\int \frac{dx}{x(\ln x+(\ln x{)}^3)}$ is equal to $\ln |\ln x|-\frac{1}{2}\ln \left(f\right(x\left)\right),$ then the value of $f\left(e^4\right)$ is:
(assume constant of integration to be zero.)

Answer 1

$I=\int \frac{dx}{x(\ln x+(\ln x{)}^3)}$
Put $\ln x=y$
$\Rightarrow \frac{dx}{x}=dy\Rightarrow I=\int \frac{dy}{y(y^2+1)}=\int \frac{dy}{y^3(1+y^{-2})}$
Put $y^{-2}+1=t$
$\Rightarrow -2y^{-3}dy=dt\Rightarrow \frac{dy}{y^3}=-\frac{dt}{2}\Rightarrow I=\int \frac{-dt}{2t}=-\frac{1}{2}\ln |t|+C=-\frac{1}{2}\ln |1+y^{-2}|(\because C=0)=-\frac{1}{2}[\ln |y^2+1|-\ln |y^2|]\Rightarrow I=-\frac{1}{2}\ln \left(\right(\ln x{)}^2+1)+\ln |\ln x|$
$\therefore$ We have, $f\left(x\right)=(\ln x{)}^2+1$
$\Rightarrow f\left(e^4\right)=16+1=17$