Question

If ${\int }_1^2{e}^{x^2}dx=a$, then ${\int }_e^{e^4}\sqrt{\ln x}dx$ is equal to

• A. $2e^4-2e-a$
• B. $2e^4-e-a$
• C. $2e^4-e-2a$
• D. $e^4-e-a$

Answer 1

Answer: (B)

$2e^4-e-a$

Let $I={\int }_e^{e^4}\sqrt{\ln x}dx$
Substitute $\sqrt{\ln x}=t\Rightarrow \frac{dx}{2x\sqrt{\ln x}}=dt\Rightarrow dx=2t{e}^{t^2}dt$
We get $I={\int }_1^{2}2t^2{e}^{t^2}dt$
Now integrating by parts, we get
$I={\left[t{e}^{t^2}\right]}_1^2-{\int }_1^2{e}^{t^2}dt$
As ${\int }_1^2{e}^{x^2}dx=a$
Therefore
$I=2e^4-e-a$
Hence, option 'B' is correct.