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.
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.
Yes, $a \log(x)$ is an antiderivative. Now evaluate it at $x=\sqrt{a}$ and at $x = a^2$, and subtract. $\frac{a \log(a)}{2} - 2 a \log(a) = -\frac{3}{2} a \log(a)$.
$a\log (x)|^{\sqrt a}_{a^2}= a\log a^{1/2}-a\log a^2={1\over 2}a\log a -2a\log a=-{3\over2}a\log a$.
Where did you get $-{1\over2}$?
$\int \mathrm1/x {d}x$ =log x now put in the values =>1/2log a -2log a =-3/2 log a