Question

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

• A. $\frac{1}{2}\ln \left(\frac{x}{\ln x}\right)-\frac{1}{4}\ln ({\ln }^{2}x-x^2)+C$
• B. $\frac{1}{4}\ln \left(\frac{\ln \left(x\right)-x}{\ln x+x}\right)-\frac{1}{2}{\tan }^{-1}\left(\frac{\ln x}{x}\right)+C$
• C. $\frac{1}{4}\ln \left(\frac{\ln \left(x\right)+x}{\ln x-x}\right)-\frac{1}{2}{\tan }^{-1}\left(\frac{\ln x}{x}\right)+C$
• D. $\frac{1}{4}(\ln \left(\frac{\ln \left(x\right)-x}{\ln x+x}\right)+{\tan }^{-1}\left(\frac{\ln x}{x}\right))+C$

Answer 1

The correct option is B $\frac{1}{4}\ln \left(\frac{\ln \left(x\right)-x}{\ln x+x}\right)-\frac{1}{2}{\tan }^{-1}\left(\frac{\ln x}{x}\right)+C$

$I=\int \frac{x^2(1-\ln x)}{{\ln }^{4}x-x^4}dx$
$=\int \frac{x^2(1-\ln x)}{x^4\left(\right(\frac{\ln x}{x}{)}^4-1)}$
$=\int \frac{1-\ln x}{x^2\left(\right(\frac{\ln x}{x}{)}^4-1)}$
Put $\frac{\ln x}{x}=t$
$\Rightarrow \left(\frac{1-\ln x}{x^2}\right)dx=dt$
$\Rightarrow I=\int \frac{dt}{t^4-1}$
$=\int \frac{1}{(t^2-1)(t^2+1)}dt$
$=\frac{1}{2}\int \frac{(t^2+1)-(t^2-1)}{(t^2-1)(t^2+1)}dt$
$=\frac{1}{2}\int \frac{1}{t^2-1}dt-\frac{1}{2}\int \frac{1}{t^2+1}dt$
$I=\frac{1}{4}\ln (\frac{t-1}{t+1})-\frac{1}{2}{\tan }^{-1}t+C$
$I=\frac{1}{4}\ln \left(\frac{\ln \left(x\right)-x}{\ln x+x}\right)-\frac{1}{2}{\tan }^{-1}\left(\frac{\ln x}{x}\right)+C$