Question

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

Answer 1

Answer: 8

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