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.
After your substitution,$$\int _1^2 \underset{D}{\underbrace{t}}\cdot \underset{I}{\underbrace {2te^{t^2}}} dt = [ te^{t^2}]_1^2 -\int_1^2 e^{t^2} dt =2e^4 -e -a$$
hint
With the substitution $$x=\sqrt{t},$$
$$\int_1^2e^{x^2}dx=\int_1^4\frac{e^t}{2\sqrt{t}}dt$$
and with
$$u=e^t$$
it becomes
$$a=\int_e^{e^4}\frac{du}{2\sqrt{\ln(u)}}$$ $$=\int_{e}^{e^4}u.\frac{du}{2u\sqrt{\ln(u)}}$$ $$=\Bigl[u\sqrt{\ln(u)}\Bigr]_{e}^{e^4}-\int_{e}^{e^4}\sqrt{\ln(u)}du$$
by parts integration. the right answer is $(D)$.
It can be found without using any substitutions at all, just with geometric meaning. Observe that for $y = e^{x^2}$ there is inverse function $x = \sqrt{\ln y}$ and for $x = 1$ and $x=2$ we have $y=e$ and $y=e^4$ (and vice versa). Check this graph in Desmos and see that $\int_{e} ^{e^4} \sqrt{\ln y} \mathrm{d}y + \int_{1}^{2} e^{x^2} \mathrm{d}x = 2\cdot e^4 - 1\cdot e$, where $2\cdot e^4$ is the area of larger rectangle and $1\cdot e$ is the area of smaller one. So, $\int_{e} ^{e^4} \sqrt{\ln x} \mathrm{d}x = 2e^4 - e - a$.
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
