What is the integral of 2x^3(1-x^4)^{-1/4} ?

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

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integral of 2x^3(1-x^4)^{-1/4}

\int 2 x^{3} (1 - x^{4})^{- \frac{1}{4}} d x

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

Solution steps

\int 2 x^{3} (1 - x^{4})^{- \frac{1}{4}} d x

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

= 2 (- \frac{1}{4} \cdot \int \frac{1}{u ^{\frac{1}{4}}} d u)

Apply the Power Rule

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

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

Substitute back

u = 1 - x^{4}

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

Simplify

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

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

Add a constant to the solution

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

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