# Integral with a dependent range

For some constant $a \in (0,1)$, Wolfram-Alpha yields

$$\int_{a^2} ^\sqrt a\frac{a}{x} \text{dx} =-\frac{3}{2} a \log(a)$$

How does one approach such an integral? I feel like the solution is probably very elementary, but I don't see it.

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Why didn't you bother asking yourself the question? Or go to sleep before posting on MSE? I feel like looking at myself when I am damn tired past 2a.m. and trying to say something intelligent while it just won't come out. – Patrick Da Silva Dec 26 '11 at 23:49
@PatrickDaSilva: See the edit, i did a simple mistake I did not find so I was confused. And yeah it was indeed a stupid question :) – Listing Dec 27 '11 at 0:06
+1 for the humility. – Patrick Da Silva Dec 27 '11 at 0:20

Yes, $a \log(x)$ is an antiderivative. Now evaluate it at $x=\sqrt{a}$ and at $x = a^2$, and subtract. $\frac{a \log(a)}{2} - 2 a \log(a) = -\frac{3}{2} a \log(a)$.

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Thank you, I will accept your answer in a few minutes :), seems like my intuition was right on this one.. – Listing Dec 26 '11 at 22:54

$a\log (x)|^{\sqrt a}_{a^2}= a\log a^{1/2}-a\log a^2={1\over 2}a\log a -2a\log a=-{3\over2}a\log a$.

Where did you get $-{1\over2}$?

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facepalm I did a mistake when simplifying the expression. Thank you – Listing Dec 26 '11 at 22:53

$\int \mathrm1/x {d}x$ =log x now put in the values =>1/2log a -2log a =-3/2 log a

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