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.

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, $\int_{e} ^{e^4} \sqrt{\ln x} \mathrm{d}x = 2e^4 - e - a$.
