What is the integral of (21x)/(\sqrt[4]{10-x^2)} ?

The integral of (21x)/(\sqrt[4]{10-x^2)} is -14(-x^2+10)^{3/4}+C

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integral of (21x)/(\sqrt[4]{10-x^2)}

- 14 (- x^{2} + 10)^{\frac{3}{4}} + C

Solution steps

Take the constant out:

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

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{4}{3} u^{\frac{3}{4}}

= 21 (- \frac{1}{2} \cdot \frac{4}{3} u^{\frac{3}{4}})

Substitute back

u = 10 - x^{2}

= 21 (- \frac{1}{2} \cdot \frac{4}{3} (10 - x^{2})^{\frac{3}{4}})

Simplify

21 (- \frac{1}{2} \cdot \frac{4}{3} (10 - x^{2})^{\frac{3}{4}}) : - 14 (- x^{2} + 10)^{\frac{3}{4}}

= - 14 (- x^{2} + 10)^{\frac{3}{4}}

Add a constant to the solution

= - 14 (- x^{2} + 10)^{\frac{3}{4}} + C

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What is the integral of (21x)/(\sqrt[4]{10-x^2)} ?
The integral of (21x)/(\sqrt[4]{10-x^2)} is -14(-x^2+10)^{3/4}+C