Question

$\int (\frac{\log x-1}{1+(\log x{)}^2}{)}^2$ is equal to

• A. $\frac{x{e}^2}{1+x^2}+C$
• B. $\frac{x}{1+(\log x{)}^2}+C$
• C. $\frac{\log x}{(\log x{)}^2+1}$
• D. None of these

Answer 1

The correct option is B $\frac{x}{1+(\log x{)}^2}+C$

$I=\int {\left(\frac{\log x-1}{1+(logx{)}^2}\right)}^{2}dx$

[ Let $logx=t[x=e^t]dx=e^{t}dt$

$\therefore I=\int {\left(\frac{t-1}{1+t^2}\right)}^2{e}^{t}dt$

$=\int \left(\frac{t^2+1-2t}{(1+t^2{)}^2}\right)e^{t}dt$

$={\int }^{e^t}(\frac{(t^2+1)}{(t^{+}1{)}^2}-\frac{2t}{(1+t^2)})dt$

$=\int e^t(\frac{1}{t^2+1}-\frac{2t}{(1+t^2{)}^2})dt$ ...............(1)

$\left[\int e^x\left(f\right(x)+f(x\left)\right)dt=e^{x}f\left(x\right)+C\right]$ ............(2)

and in equation (1) we see that

if $f\left(x\right)=\frac{1}{t^2+1}$ then $f^1\left(x\right)=\frac{-2t}{(t^2+1{)}^2}$

$\therefore$ from equation (1) & equation (2)

$I=e^t\left(\frac{1}{t^2+1}\right)+C=e^{logx}\left(\frac{1}{(logx{)}^2+1}\right)+C$

$I=x(\frac{1}{(logx{)}^2}+1)+C$ as So, option (B) is correct