Question

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

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

Answer 1

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

Given ${\int }_1^2{e}^{x^2}dx=a,$

Let $I={\int }_e^{e^4}\sqrt{\ln \left(x\right)}dx$

Put $\ln \left(x\right)=t^2\Rightarrow \frac{1}{x}dx=2tdt$

$\therefore I={\int }_1^2{e}^{t^2}.{2t}^{2}dt={\left(t{e}^{t^2}\right)}_1^2-{\int }_1^2{e}^{t^2}dt={2e}^4-e-a.$