Fairly simple but incorrect integral

[3.141592654]

Homework Statement

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

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

Homework Equations

The Attempt at a Solution

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

$$u=\sqrt[3]{x-2}$$

$$u^3=x-2$$

$$3udu=dx$$

$$=3\int\frac{udu}{1+u}$$

$$w=1+u$$

$$w-1=u$$

$$dw=du$$

$$=3\int\frac{w-1dw}{w}$$

$$=3\int\frac{wdw}{w}-3\int\frac{dw}{w}$$

$$=3\int dw-3\int\frac{dw}{w}$$

$$=3w-3ln|w|+C$$

$$=3(1+u)-3ln|1+u|+C$$

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

$$=3+3(x-2)^\frac{1}{3}-3ln|1+(x-2)^\frac{1}{3}|+C$$

 

[Mark44]

Your equation for dx is wrong. If u³ = x -2, then 3u²du = dx.