# Definite Integral

## Homework Statement

$$\int_0^a {x\sqrt{x^2+a^2}\,dx}$$

Also, (A>0)

## The Attempt at a Solution

Firstly, I set

u=x^2+a^2

Then take the derivative,

du=2x dx

$$1/2\int_0^a {\sqrt{u}\,du}$$

Now I integrated. So

(1/3) * [(x^2+a^2)^3/2] from a to 0.

I ended up with

(1/3)[(a^2+a^2)^(3/2)-a^3]

This is where I get lost. It must have something to do with the (A>0). The answer in the book is:

(1/3)(2*sqrt(2)-1)a^3

I can't see how to eliminate the a's to get the (2*sqrt(2)-1).

Thanks for the help.

Related Calculus and Beyond Homework Help News on Phys.org
$$\frac 1 3[(a^2+a^2)^\frac{3}{2}-a^3]$$

$$\frac 1 3[(2a^2)^\frac{3}{2}-a^3]$$

$$\frac 1 3[2^\frac{3}{2}a^2^\frac{3}{2}-a^3)]$$

Continue simplifying.