Question

If ${\int }_1^2{e}^{x^2}dx=a$,then the value of ${\int }_e^{e^4}\sqrt{ln\left(x\right)}dx$ is:

• A. $e^4-e$
• B. $e^4-a$
• C. $3e^4-a$
• D. $2e^4-e-a$

Answer 1

The correct option is D $2e^4-e-a$

Given:

${\int }_1^2{e}^{x^2}dx=a$ $\left(1\right)$

Let $S={\int }_e^{e^4}\sqrt{\ln x}dx$

Substituting $e^{x^2}=t$ in Eq. $\left(1\right)$:

$e^{x^2}=t$

$x=\sqrt{\ln t}$

$d\left(e^{x^2}\right)=dt$

$e^{x^2}dx=\frac{dt}{2x}$

$\therefore a={\int }_e^{e^4}\frac{dt}{2\sqrt{\ln t}}$

$S={\int }_e^{e^4}\sqrt{\ln x}dx$

Applting Integration by parts:

$S=e^4\sqrt{\ln e^4}-e\sqrt{\ln e}-{\int }_e^{e^4}\left(\frac{t}{2\sqrt{\ln t}}\frac{1}{t}dt\right)$

$\therefore S=2e^4-e-a$

Hence, option D is correct.