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.