Different way to do this integral

[iRaid]

Homework Statement

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

Homework Equations

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
Therefore the answer is:
\frac{3(x^{2}+1)^{\frac{5}{3}}}{5}-\frac{3(x^{2}+1)^{\frac{2}{3}}}{2}+C

 

[Dick]

Homework Statement

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

Homework Equations

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
Therefore the answer is:
\frac{3(x^{2}+1)^{\frac{5}{3}}}{5}-\frac{3(x^{2}+1)^{\frac{2}{3}}}{2}+C

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

 

[iRaid]

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

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?

 

[Dick]

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?

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

 

[iRaid]

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

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

Idk what I'd do with that :\

 

[Dick]

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

Idk what I'd do with that :\

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

 

[iRaid]

Am I going about this the right way:
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

 

[Dick]

Am I going about this the right way:
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

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

 

[iRaid]

Looks ok to me.

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 :)