Why Is This Integral Challenging?

[PhysicsMark]

Homework Statement

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.

Homework Equations

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

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?

 

[Dick]

u=1+x^2 will do it.

 

[jgens]

The substitution x=tan(u) also works.

 

[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.

 

[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.