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

The integral of (10x^3-5x)\sqrt[5]{x^4-x^2} is 25/12 (x^4-x^2)^{6/5}+C

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integral of (10x^3-5x)\sqrt[5]{x^4-x^2}

\int (10 x^{3} - 5 x) 5 x^{4} - x^{2} d x

\frac{25}{12} (x^{4} - x^{2})^{\frac{6}{5}} + C

Solution steps

\int (10 x^{3} - 5 x) 5 x^{4} - x^{2} d x

Apply u-substitution

: \int \frac{5 5 u}{2} d u

= \int \frac{5 5 u}{2} d u

Take the constant out:

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

= \frac{5}{2} \cdot \int 5 u d u

Apply the Power Rule

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

= \frac{5}{2} \cdot \frac{5}{6} u^{\frac{6}{5}}

Substitute back

u = x^{4} - x^{2}

= \frac{5}{2} \cdot \frac{5}{6} (x^{4} - x^{2})^{\frac{6}{5}}

Simplify

\frac{5}{2} \cdot \frac{5}{6} (x^{4} - x^{2})^{\frac{6}{5}} : \frac{25}{12} (x^{4} - x^{2})^{\frac{6}{5}}

= \frac{25}{12} (x^{4} - x^{2})^{\frac{6}{5}}

Add a constant to the solution

= \frac{25}{12} (x^{4} - x^{2})^{\frac{6}{5}} + C

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What is the integral of (10x^3-5x)\sqrt[5]{x^4-x^2} ?
The integral of (10x^3-5x)\sqrt[5]{x^4-x^2} is 25/12 (x^4-x^2)^{6/5}+C