# Homework Help: Different way to do this integral

1. Feb 10, 2013

### iRaid

1. The problem statement, all variables and given/known data
Make a substitution to express the integrand as a rational
function and then evaluate the integral.
$$\int \frac{x^{3}}{\sqrt[3]{x^{2}+1}}dx$$

2. Relevant equations

3. The attempt at a solution
I didn't know what to substitute using their directions, but I got the answer (I think), could someone tell me if my answer is right and how to do it using the directions.
$$u=x^{2} \\du=2xdx$$
$$\int \frac{u}{\sqrt[3]{u+1}} du$$
$$v=u+1 \\dv=du$$
$$\int \frac{v-1}{\sqrt[3]{v}} dv = \int v^{\frac{2}{3}}-v^{\frac{-1}{3}}dv$$
$$\frac{3(x^{2}+1)^{\frac{5}{3}}}{5}-\frac{3(x^{2}+1)^{\frac{2}{3}}}{2}+C$$

2. Feb 10, 2013

### Dick

I think you mislaid a 1/2 factor. But why not substitute u=x^2+1 to begin with?

3. Feb 10, 2013

### iRaid

Yeah that works too, and I did forget the 1/2. Yeah I could do the u=x^2+1 too, idk what I was thinking lol. But is there any other way to do it using the directions?

4. Feb 10, 2013

### Dick

If you want to rationalize it completely try u^3=x^2+1.

5. Feb 11, 2013

### iRaid

What does du become then?
$$du=\frac{2}{3}\frac{x}{(x^2+1)^{\frac{2}{3}}}$$?

Idk what I'd do with that :\

6. Feb 11, 2013

### Dick

3u^2du=2xdx. The usual routine. Just keep going.

7. Feb 11, 2013

### iRaid

$$u^{3}=x^{2}+1\\3u^{2}du=2xdx\\\frac{3}{2} \int \frac{(u^{3}-1)(u^{2})}{u}du$$
=$$\int u^{4}-udu$$

8. Feb 11, 2013

### Dick

Looks ok to me. In terms of x it's the same answer you got before, right?

9. Feb 11, 2013

### iRaid

Seems like a much easier way to do this. I didn't know you could take the derivative of each side of the u-substitution like that also.

Thanks for the help :)