What is the integral of (x^3)/(sqrt(1-x^4)) ?

The integral of (x^3)/(sqrt(1-x^4)) is -1/2 sqrt(1-x^4)+C

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integral of (x^3)/(sqrt(1-x^4))

\int \frac{x ^{3}}{1 - x ^{4}} d x

- \frac{1}{2} 1 - x^{4} + C

Solution steps

\int \frac{x ^{3}}{1 - x ^{4}} d x

Apply u-substitution

: \int - \frac{1}{4 u} d u

= \int - \frac{1}{4 u} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

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

Substitute back

u = 1 - x^{4}

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

Simplify

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

= - \frac{1}{2} 1 - x^{4}

Add a constant to the solution

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

f (x) = 5 - 3 x

\int_{- 1}^{1} (x^{2} - x) d x

\int 4 r d r

\int \frac{1}{x - 9} d x

\frac{y}{x - 5 y} = x^{7} + 9

What is the integral of (x^3)/(sqrt(1-x^4)) ?
The integral of (x^3)/(sqrt(1-x^4)) is -1/2 sqrt(1-x^4)+C