66Let
F(a)=\int_a^{a^2}\dfrac{\mathrm dx}{x+\sqrt{x}}
Setting F'(a)=0, we get
\dfrac{2a}{a^2+a}-\dfrac{1}{a+\sqrt{a}}=0
Since a>0, this gives
\dfrac{2}{a+1}=\dfrac{1}{a+\sqrt{a}}
which has the only solution a=3-2\sqrt{2}=(\sqrt{2}-1)^2 in positive reals.
It is easy to see that this value of a corresponds to a minimum and is same as tan2(Ï€/8).
