# Find $\int_{e}^{e^4}\sqrt{\ln x}~dx$

If $$\int_{1}^{2}e^{x^2}~dx=\alpha$$ ,then show that $$\int_{e}^{e^4}\sqrt{\ln x}~dx=2e^4-e-\alpha$$

How I proceed:$$\int_{e}^{e^4}\sqrt{\ln x}~dx=2\int_{1}^{2}z^2e^{z^2}~dx$$ Then I can't proceed. Please help.

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The first integral has been edited to change $e^{x^2}$ to $e^2$. –  Daryl Dec 20 '12 at 10:56
Integrating by parts we get $$\int _1 ^2 z \cdot 2z e^{z^2} \, dz = \Big [ze^{z^2} \Big ]_1 ^2 - \int _1 ^2 e^{z^2}\, dz.$$
You can also see it pictorially. In the figure below, the curve between the red and blue areas is $y=e^{x^2}$, so the blue area is $\alpha$, and the desired integral is the red area. The area of the large rectangle is clearly $2e^4$, and the area of the black rectangle is $e$, so $$\int_{e}^{e^4}\sqrt{\ln x}~dx=\text{red area}=2e^4-e-\text{blue area}=2e^4-e-\alpha\;$$