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!
Best Answer
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.