Question

$\int {\left\{\frac{logx-1}{1+{\left(logx\right)}^2+1}\right\}}^2$ dx is equal to

• A. $\frac{logx}{{\left(logx\right)}^2+1}+c$
• B. $\frac{x}{x^2+1}+c$
• C. $\frac{x{e}^x}{1+x^2}+c$
• D. $\frac{x}{{\left(logx\right)}^2+1}+c$

Answer 1

Answer: (D)

$\int \frac{{\left(logx-1\right)}^2}{{\left(1+{\left(logx\right)}^2\right)}^2}dx$
$=\int \left[\frac{1}{{\left(1+\left(logx\right)\right)}^2}-\frac{2logx}{{\left(1+{\left(logx\right)}^2\right)}^2}\right]dx$
$=\int \left[\frac{e^t}{1+t^2}-\frac{2t{e}^t}{{\left(1+t^2\right)}^2}\right]dt$
( Putting $logx=t\Rightarrow dx=e^{t}dt)$
$=\int e^t\left[\frac{1}{1+t^2}-\frac{2t}{{\left(1+t^2\right)}^2}\right]dt$
$=\frac{e^t}{1+t^2}+c$
$=\frac{x}{1+{\left(logx\right)}^2}+c$(Resubstituting $t=logx)$