Let $\int_{1}^{2} e^{x^2} \mathrm{d}x=a$. Then find the value of $\int_{e} ^{e^4} \sqrt{\ln x} \mathrm{d}x$ Let $\int_{1}^{2} e^{x^2} \mathrm{d}x =a$. Then the value of $\int_{e} ^{e^4} \sqrt{\ln x} \mathrm{d}x $ is
(A) $e^4-a$
(B) $2e^4 - a$
(C)$e^4 - e - 4a$
(D) $2e^4-e-a$
$\bf{Try:}$
Let $\ln x =t^2$ then $x=e^{t^2}$ and $dx=2te^{t^2} dt$. Then the given integral changes to $\int_{1}^{2} 2t^2 e^{t^2}dt$ but $\int 2t^2 e^{t^2}dt=t^2\int e^{t^2} dt - \int \left(2t\int e^{t^2}dt\right)dt$
I'm unable to progress further. Please help thanks in advance.

Define I = $∫_1^2\ exp(x^2)dx $ and $I_1 = \int_{e} ^{e^4} \sqrt{\ln x} \mathrm{d}x$
Let u= exp ($x^2$),
then du = 2xexp($x^2$) dx.
you already show
$I_1 = ∫_1^22t^2 exp(t^2)dt $
so that $I_1$ = $∫_1^2\ xdu $.
Notice my x is your t.
Using integration by parts, we have
$I_1$ = $xu|_{x=2} -xu|_{x=1} - ∫_0^1 u dx$
= $x exp(x^2)|x=2 - xexp(x^2)|x=1 -∫_1^2{exp(x^2)} dx$
= 2$e^4$ - e - I
= 2$e^4$ - e - a by definition of a
