Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 2 down vote accepted

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

share|cite|improve this answer
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 $-{1\over2}$?

share|cite|improve this answer
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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.