Originally Posted by
topsquark $\displaystyle \int_0^1 \frac{dx}{3+\sqrt x}$
Now set:
$\displaystyle t = \sqrt x $
Thus
$\displaystyle dt = \frac{1}{2} \frac{1}{\sqrt x} dx$
or
$\displaystyle dx = 2 \sqrt x dt = 2 t \, dt$
For the limits on the integral:
Upper limit: $\displaystyle t = \sqrt 1 = 1$
Lower limit: $\displaystyle t = \sqrt 0 = 0$
Thus:
$\displaystyle \int_0^1 \frac{dx}{3+\sqrt x}$
$\displaystyle = \int_0^1 \frac{2 t \, dt}{3 + t} =2 \int_0^1 dt \frac{t}{t+3}=2 \int_0^1 dt \left ( 1 - \frac{3}{t+3} \right )$
$\displaystyle =2t|_0^1 - 6 \, ln|t+3| |_0^1 = 2 - 6 \, ln4 + 6 \, ln 3$
$\displaystyle = 2 + 6 \, ln \left ( \frac{3}{4} \right ) $
-Dan