# Simple Integral

1. Sep 1, 2010

### PhysicsMark

1. The problem statement, all variables and given/known data

Hello. I have a simple integral here that has been stumping me for the last 30 minutes. It appears that my basic integration skills have gotten very rusty.

2. Relevant equations

$$\int{x^3}\sqrt{1+x^2}dx$$

3. The attempt at a solution
I am pretty sure a simple substitution will do, but I have yet to find it. I have tried simplifying the expression various ways, integration by parts, and I have also tried a few substitutions. Any ideas?

2. Sep 1, 2010

### Dick

u=1+x^2 will do it.

3. Sep 1, 2010

### jgens

The substitution x=tan(u) also works.

4. Sep 1, 2010

### PhysicsMark

Thank you Dick and Jgens.

$$\int{x^3}\sqrt{1+x^2}dx$$

Let u=1+x^2$$\rightarrow$$du=2xdx

$$\int{x^3}\sqrt{1+x^2}dx=\frac{1}{2}\int(u-1)\sqrt{u}du$$

$$\frac{1}{2}\int(u-1)\sqrt{u}du=\frac{1}{2}(\int{u^{\frac{3}{2}}}du-\int{u^{\frac{1}{2}}du)$$

$$\int{u^{\frac{3}{2}}}du=\frac{2}{5}u^{\frac{5}{2}}$$

$$\int{u^{\frac{1}{2}}du=\frac{2}{3}u^{\frac{3}{2}}$$

$$\frac{1}{2}\int(u-1)\sqrt{u}du=\frac{2}{10}u^{\frac{5}{2}}-\frac{2}{6}u^{\frac{3}{2}}$$

It appears I made a mistake somewhere. I believe my result should be multiplied by 1/2, and I can't find where I left that out.

5. Sep 1, 2010

### PhysicsMark

Nevermind. I guess the second part of the problem is incorrect. Mathematica has arrived at the same conclusion as this also. Thanks for all the help.