Question

The value of ${\int }_1^e{\left[\frac{\log x-1}{1+(\log x{)}^2}\right]}^{2}dx$ is equal to

• A. $\frac{1}{2}(e-2)$
• B. $\frac{(e-1{)}^2}{2(1+e^2)}$
• C. $\frac{1}{2}(e-1)$
• D. $\frac{1}{2}e$

Answer 1

The correct option is A $\frac{1}{2}(e-2)$

${\int }_1^e\frac{(\log x-1{)}^2}{(1+(\log x{)}^2{)}^2}dx={\int }_1^e[\frac{1}{(1+(\log x{)}^2)}-\frac{2\log x}{(1+(\log x{)}^2{)}^2}]dx$
Substitute $\log x=t\Rightarrow dx=e^{t}dt$
$={\int }_0^1[\frac{e^t}{1+t^2}-\frac{2t{e}^t}{(1+t^2{)}^2}]dt$
$={\int }_0^1{e}^t[\frac{1}{1+t^2}-\frac{2t}{(1+t^2{)}^2}]dt$
$={\left[\frac{e^t}{1+t^2}\right]}_0^1=\frac{e}{2}-1=\frac{1}{2}(e-2)$