Evaluate: $\int\limits_{e}^{e^4} \sqrt{\ln x} dx$ 
Question:

Given
$$\int_1^2 e^{x^2} dx= a$$
Evaluate:
$$\int_{e}^{e^4} \sqrt{\ln x} dx$$

Answer:
$$2e^4-e-a$$

My Attempt:
I began with the substitution $\sqrt{\ln x}=t$, which transforms the required integral (say, $I$) to:
$$I=2\int_1^2 t^2e^{t^2} dt$$
Since the limits match with the "known" integral's limits, I thought of applying "By Parts" twice to eliminate the unwanted $t^2$ that's lurking. But I'm stuck:
$$I=\left[t^2\left(\int e^{t^2}dt\right)\right]_1^2-\int_1^2(2t)\left(\int e^{t^2}dt\right)dt$$
Now I'm unable to proceed further because:
A) Evaluation of
$$\int e^{x^2}dx$$
is not in our syllabus.
B) "Interchanging" the functions in the formula is not an option as that will lead us away from the destination.
Any help would be great!
 A: You had the right idea for integration by parts, but you should chose your $u$ and $dv$ slightly differently.  If you let $dv=te^{t^2} \; dt$ and $u=t$ then it will work.  Note that $v$ can be evaluated by a simple substitution of $\phi=t^2$.
\begin{align}
I&=te^{t^2} \big \rvert_1^2 - \int_1^2 e^{t^2} \; dt \\
&=\boxed{2e^4-e-a}
\end{align}
Where $\int_1^2 e^{t^2} \; dt=a$ as stated in the question.
A: You can do this geometrically.  The functions $e^{x^2}$ and $\sqrt{\ln x}$ are inverses of each other.  Sketch the bit of curve $y=e^{x^2}$ from $x=1$ to $x=2$.  The top right point of the curve is $(2,e^4)$ and the bottom left point is $(1,e)$.
The rectangle $[0,2]\times[0,e^4]$ is cut into $3$ pieces.  A small rectangle $[0,1]\times [0,e]$, and two pieces which are the two integrals in this problem.  The area you want is from $y=e$ to $y=e^4$ and between the $y$-axis and the curve.  So it's
$$\mbox{Area of big rectangle } - \mbox{ Area of little rectangle } - \mbox{ Area from given integral }$$
$$=2e^4 -e -a.$$

A: note that $e^{x^2}$ and $\sqrt{\ln x}$ are inverse functions
Also it can be easily shown that $$\int_{a}^bf(x)dx+\int_{f(a)}^{f(b)}f^{-1}(x)dx=bf(b)-af(a)$$
the rest can be done easily
