Question

For $x>0$ and arbitrary constant of integration $C,$ $\int \frac{x^2(1-\ln x)}{(\ln x{)}^4-x^4}dx$ equals

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

Answer 1

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

$I=\int \frac{x^2(1-\ln x)}{x^4({\left(\frac{\ln x}{x}\right)}^4-1)}dx=\int \frac{1-\ln x}{x^2({\left(\frac{\ln x}{x}\right)}^4-1)}dx$

Put $\frac{\ln x}{x}=t\Rightarrow \frac{1-\ln x}{x^2}dx=dt$

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