Question

If $f(x)=\int \frac{3(\ln x)^{2}+(\ln x)^{4}}{x\left(1+(\ln x)^{2}-(\ln x)^{3}\right)^{2}} d x$ and $f(1)=0$, then the value of $\left|3 f\left(e^{2}\right)\right|$ is equal to

Answer 1

Put $\ln x=t \Rightarrow d x=e^{t} d t$
$=\int \frac{3 t^{2}+t^{4}}{\left(1+t^{2}-t^{3}\right)^{2}} d t$
$=\int \frac{\left(3 t^{2}+t^{4}\right)}{t^{6}\left(\frac{1}{t^{3}}+\frac{1}{t}-1\right)^{2}} d t$
$=\int \frac{\frac{3}{t^{4}}+\frac{1}{t^{2}}}{\left(\underbrace{\frac{1}{t^{3}}+\frac{1}{t}-1}_{z}\right)} d t=-\int \frac{d z}{z^{2}}=\frac{1}{z}+C$
$\Rightarrow f(x)=\frac{(\ln x)^{3}}{1+(\ln x)^{2}-(\ln x)^{3}}+C$
$\therefore f(x)=0+C$
$\Rightarrow C=0$
$\therefore f(x)=\frac{(\ln x)^{2}}{1+(\ln x)^{2}-(\ln x)^{3}}$
$\Rightarrow 3 f\left( e ^{2}\right)=-8$