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.