Answer
$\int_{0}^{a}x~\sqrt{x^2+a^2}~dx = \frac{1}{3}~(2\sqrt{2}-1)~a^3$
Work Step by Step
$\int_{0}^{a}x~\sqrt{x^2+a^2}~dx$
Let $u = x^2+a^2$
$\frac{du}{dx} = 2x$
$dx = \frac{du}{2x}$
When $x = 0$, then $u =a^2$
When $x = a$, then $u = 2a^2$
$\int_{a^2}^{2a^2} x~\sqrt{u}~\frac{du}{2x}$
$=\int_{a^2}^{2a^2} \frac{1}{2}\cdot \sqrt{u}~du$
$= \frac{1}{2}~(\frac{2}{3}u^{3/2}~\vert_{a^2}^{2a^2})$
$= \frac{1}{3}~(u^{3/2}~\vert_{a^2}^{2a^2})$
$= \frac{1}{3}~[~(2a^2)^{3/2}-(a^2)^{3/2}~]$
$= \frac{1}{3}~(2\sqrt{2}~a^3-a^3)$
$= \frac{1}{3}~(2\sqrt{2}-1)~a^3$
