Question

$\int \frac{x^2(1-\ln x)}{{ln}^{4}x-x^4}dx$ equals

Answer 1

$\int \frac{n^2\left(\ln n\right)}{(\ln n{)}^4-n^4}dx$

$\Rightarrow \int \frac{(1-\ln n)}{\frac{(\ln n{)}^4-n^4}{n^2}}dx$

$\Rightarrow \int \frac{(1-\ln n)}{n^2({\left(\frac{\ln n}{n}\right)}^4-1)}dx$

Put $\frac{\ln n}{n}=t$

$(-\ln n\left(\frac{1}{n^2}\right)+\frac{1}{n^2})dx=dt$

$\Rightarrow \int \frac{dt}{t^4-1}$

$\Rightarrow \frac{1}{2}\int [\frac{1}{t^2-1}-\frac{1}{t^2+1}]dx$

$\Rightarrow \frac{1}{2}\int \frac{dt}{t^2-1}-\frac{1}{2}\int \frac{dt}{t^2+1}$

Using formula $\int \frac{1}{n^2-n^2}dx=\frac{1}{2a}\ln \left(\frac{n-a}{n+a}\right)$

$\int \frac{1}{a^2+n^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{n}{a}\right)$

$\Rightarrow \frac{1}{2}\left[\frac{1}{2}\ln \left(\frac{t-1}{t+1}\right)\right]\frac{-1}{2}.{\tan }^{-1}t$.

$\Rightarrow \frac{1}{4}\ln \left(\frac{t-1}{t+1}\right)-\frac{1}{2}{\tan }^{-1}t$

$t=\frac{\ln n}{n}$

So,

$\Rightarrow \frac{1}{4}\ln \left(\frac{\ln n-1}{\ln n+1}\right)-\frac{1}{2}{\tan }^{-1}\left(\frac{\ln n}{n}\right)$