What is the integral of x\sqrt[4]{1-x^2} ?

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

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integral of x\sqrt[4]{1-x^2}

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

= - \frac{1}{2} \cdot \frac{4}{5} u^{\frac{5}{4}}

Substitute back

u = 1 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int t \cdot e^{- t} d t

(y + y cos (x y)) d x + (x + x cos (x y)) d y = 0

\int_{1}^{e} ln (9 x) d x

\frac{\partial}{\partial x} (x^{5} e^{x y} + 3 x)

\frac{3}{8 x ^{\frac{5}{2}}}

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