What is the integral of x^2\sqrt[5]{x^3+1} ?

The integral of x^2\sqrt[5]{x^3+1} is 5/18 (x^3+1)^{6/5}+C

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integral of x^2\sqrt[5]{x^3+1}

\frac{5}{18} (x^{3} + 1)^{\frac{6}{5}} + 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 u-substitution

Apply the Power Rule

: \frac{5}{6} v^{\frac{6}{5}}

= \frac{1}{3} \cdot \frac{5}{6} v^{\frac{6}{5}}

Substitute back

= \frac{1}{3} \cdot \frac{5}{6} (x^{3} + 1)^{\frac{6}{5}}

Simplify

\frac{1}{3} \cdot \frac{5}{6} (x^{3} + 1)^{\frac{6}{5}} : \frac{5}{18} (x^{3} + 1)^{\frac{6}{5}}

= \frac{5}{18} (x^{3} + 1)^{\frac{6}{5}}

Add a constant to the solution

= \frac{5}{18} (x^{3} + 1)^{\frac{6}{5}} + C

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What is the integral of x^2\sqrt[5]{x^3+1} ?
The integral of x^2\sqrt[5]{x^3+1} is 5/18 (x^3+1)^{6/5}+C