What is the integral of (x^4)/(\sqrt[3]{x^5-2)} ?

The integral of (x^4)/(\sqrt[3]{x^5-2)} is 3/10 (x^5-2)^{2/3}+C

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integral of (x^4)/(\sqrt[3]{x^5-2)}

\frac{3}{10} (x^{5} - 2)^{\frac{2}{3}} + C

Solution steps

Apply u-substitution

Take the constant out:

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

Apply the Power Rule

: \frac{3}{2} u^{\frac{2}{3}}

= \frac{1}{5} \cdot \frac{3}{2} u^{\frac{2}{3}}

Substitute back

u = x^{5} - 2

= \frac{1}{5} \cdot \frac{3}{2} (x^{5} - 2)^{\frac{2}{3}}

Simplify

\frac{1}{5} \cdot \frac{3}{2} (x^{5} - 2)^{\frac{2}{3}} : \frac{3}{10} (x^{5} - 2)^{\frac{2}{3}}

= \frac{3}{10} (x^{5} - 2)^{\frac{2}{3}}

Add a constant to the solution

= \frac{3}{10} (x^{5} - 2)^{\frac{2}{3}} + C

\int \frac{7}{( 1 - 6 x ) ^{6}} d x

\int - \frac{x ^{2}}{2 ( 5 - x )} d x

\int \frac{x + 3}{x ^{2} - 4} d x

\int \frac{1}{x ^{2} + 9 x} d x

\int \frac{x}{( 4 + x ^{2} ) ^{\frac{3}{2}}} d x

What is the integral of (x^4)/(\sqrt[3]{x^5-2)} ?
The integral of (x^4)/(\sqrt[3]{x^5-2)} is 3/10 (x^5-2)^{2/3}+C