What is the integral of (x^2)/(\sqrt[3]{2-x)} ?

The integral of (x^2)/(\sqrt[3]{2-x)} is -3/8 (2-x)^{8/3}+12/5 (2-x)^{5/3}-6(2-x)^{2/3}+C

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integral of (x^2)/(\sqrt[3]{2-x)}

\int \frac{x ^{2}}{3 2 - x} d x

- \frac{3}{8} (2 - x)^{\frac{8}{3}} + \frac{12}{5} (2 - x)^{\frac{5}{3}} - 6 (2 - x)^{\frac{2}{3}} + C

Solution steps

\int \frac{x ^{2}}{3 2 - x} d x

Apply u-substitution

= \int - \frac{( - u + 2 ) ^{2}}{3 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{( - u + 2 ) ^{2}}{3 u} d u

Expand

\frac{( - u + 2 ) ^{2}}{3 u} : u^{\frac{5}{3}} - 4 u^{\frac{2}{3}} + \frac{4}{3 u}

= - \int u^{\frac{5}{3}} - 4 u^{\frac{2}{3}} + \frac{4}{3 u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (\int u^{\frac{5}{3}} d u - \int 4 u^{\frac{2}{3}} d u + \int \frac{4}{3 u} d u)

\int u^{\frac{5}{3}} d u = \frac{3}{8} u^{\frac{8}{3}}

\int 4 u^{\frac{2}{3}} d u = \frac{12}{5} u^{\frac{5}{3}}

\int \frac{4}{3 u} d u = 6 u^{\frac{2}{3}}

= - (\frac{3}{8} u^{\frac{8}{3}} - \frac{12}{5} u^{\frac{5}{3}} + 6 u^{\frac{2}{3}})

Substitute back

u = 2 - x

= - (\frac{3}{8} (2 - x)^{\frac{8}{3}} - \frac{12}{5} (2 - x)^{\frac{5}{3}} + 6 (2 - x)^{\frac{2}{3}})

Simplify

= - \frac{3}{8} (2 - x)^{\frac{8}{3}} + \frac{12}{5} (2 - x)^{\frac{5}{3}} - 6 (2 - x)^{\frac{2}{3}}

Add a constant to the solution

= - \frac{3}{8} (2 - x)^{\frac{8}{3}} + \frac{12}{5} (2 - x)^{\frac{5}{3}} - 6 (2 - x)^{\frac{2}{3}} + C

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What is the integral of (x^2)/(\sqrt[3]{2-x)} ?
The integral of (x^2)/(\sqrt[3]{2-x)} is -3/8 (2-x)^{8/3}+12/5 (2-x)^{5/3}-6(2-x)^{2/3}+C