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

The integral of x/(\sqrt[4]{5-2x^2)} is -1/3 (5-2x^2)^{3/4}+C

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

- \frac{1}{3} (5 - 2 x^{2})^{\frac{3}{4}} + 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{4}{3} u^{\frac{3}{4}}

= - \frac{1}{4} \cdot \frac{4}{3} u^{\frac{3}{4}}

Substitute back

u = 5 - 2 x^{2}

= - \frac{1}{4} \cdot \frac{4}{3} (5 - 2 x^{2})^{\frac{3}{4}}

Simplify

- \frac{1}{4} \cdot \frac{4}{3} (5 - 2 x^{2})^{\frac{3}{4}} : - \frac{1}{3} (5 - 2 x^{2})^{\frac{3}{4}}

= - \frac{1}{3} (5 - 2 x^{2})^{\frac{3}{4}}

Add a constant to the solution

= - \frac{1}{3} (5 - 2 x^{2})^{\frac{3}{4}} + C

e^{p}

\frac{3}{5 ( s - 3 ) ^{2} + 80}

x^{2} + 2 x + 1, 2 x + 5, - 2, 2

\int x^{2} + 4 x + 5 d x

y d y = 4 x y^{2} + 1 d x

What is the integral of x/(\sqrt[4]{5-2x^2)} ?
The integral of x/(\sqrt[4]{5-2x^2)} is -1/3 (5-2x^2)^{3/4}+C