Solve Integral: \int\frac{dx}{1+\sqrt[3]{x-2}}

[3.141592654]

Homework Statement

\int\frac{dx}{1+\sqrt[3]{x-2}}

Homework Equations

The answer is given: \frac{3}{2}(x-2)^\frac{3}{2}-3(x-2)^\frac{1}{3}+ln|1+(x-2)^\frac{1}{3}|+C

I have to get my answer to look just like this.

The Attempt at a Solution

u=x-2, du=dx

=\int\frac{dx}{1+\sqrt[3]{u}}

w=1+\sqrt[3]{u}

dw=\frac{du}{3u^\frac{2}{3}}

3u^\frac{2}{3}dw=du

(w-1)^2=u^\frac{2}{3}

3(w-1)^2dw=du

=3\int\frac{(w-1)^2dw}{w}

=3\int\frac{w^2-2w+1}{w}dw

=3\int\frac{w^2}{w}dw-3\int\frac{2w}{w}dw+3\int\frac{1}{w}dw

=3\int\(wdw-6\int\(dw+3\int\frac{dw}{w}

=\frac{3}{2}(w^2)-6w+3ln|w|+C

=\frac{3}{2}(1+\sqrt[3]{u})^2-6(1+\sqrt[3]{u})+3ln|1+\sqrt[3]{u}|+C

=\frac{3}{2}(1+\sqrt[3]{x-2})^2-6(1+\sqrt[3]{x-2})+3ln|1+\sqrt[3]{x-2}|+C

From here I don't know where to go to get the answer stated above or if I'm not on the right track.

 

[Dick]

You are doing fine. Now, just multiply the first two terms out. You can throw the constants out, since you already have a '+C'. Now combine what's left.