Computing the definite integral $\int _0^a \:x \sqrt{x^2+a^2} \,\mathrm d x$ 
Compute the following definite integral $$\int _0^a \:x \sqrt{x^2+a^2} \,\mathrm d x$$

This is what I did:
$u = x^2 + a^2 $
$du/dx = 2x$
$du = 2xdx$
$1/2 du = x dx$
$\int _0^a\:\frac{1}{2}\sqrt{u}du = \frac{1}{2}\cdot \frac{u^{\frac{3}{2}}}{\left(\frac{3}{2}\right)}$  from $0$ to $a$.
$\frac{1}{3}\cdot \left(x^2+a^2\right)^{\frac{3}{2}}$ from $0$ to $a$.
I eventually got:
$\frac{1}{3}\left(81+a^2\right)^{\frac{3}{2}}-\frac{1}{3}\left(a^2\right)^{\frac{3}{2}}$
but this was incorrect.
The correct answer was:
$\frac{1}{3}\left(2\sqrt{2}-1\right)a^3$
Any help?
 A: Hint
You made mistakes here:
$$\frac{1}{3}\cdot \left(x^2+a^2\right)^{\frac{3}{2}}\big |_0^a=\frac 1 3 (2a^2)^{\frac 3 2}-\frac {a^3}{3}=\frac 1 3 (\sqrt{2^3}|a|^3)-\frac {a^3}{3}=....$$
You were almost done...
A: You calculated the antiderivative correctly. 
$$\frac{1}{3}\cdot \left(x^2+a^2\right)^{\frac{3}{2}}$$ from $0$ to $a$ is $$\frac{1}{3}\cdot \left(a^2+a^2\right)^{\frac{3}{2}}-\frac{1}{3}\cdot \left(0^2+a^2\right)^{\frac{3}{2}}=\\\frac{1}{3}\cdot((2a^2)^\frac{3}{2}-(a^2)^\frac{3}{2})=\frac{1}{3}\cdot(2^\frac{3}{2}a^3-a^3)=\frac{1}{3}(2^\frac{3}{2}-1)a^3$$
Which explains the answer in your book. I don't know how you got $$\frac{1}{3}\left(81+a^2\right)^{\frac{3}{2}}-\frac{1}{3}\left(a^2\right)^{\frac{3}{2}}$$ from the previous step.
A: You can simply use the reverse chain rule for integration and you get:
$$\int_0^a x\sqrt{x^2+a^2}dx=\frac 12\int_0^a 2x\sqrt{x^2+a^2}dx=\left[\frac{(x^2+a^2)^\frac 32}3\right]_0^a=\\=\frac{(2a^2)^\frac 32}3-\frac{(a^2)^\frac 32}3=\color{red}{\frac{(2\sqrt 2-1)}3|a|^3}$$
