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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u - 1}{32 u ^{\frac{3}{4}}} d u

= 2 \cdot \int \frac{u - 1}{32 u ^{\frac{3}{4}}} d u

Take the constant out:

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

= 2 \cdot \frac{1}{32} \cdot \int \frac{u - 1}{u ^{\frac{3}{4}}} d u

Expand

\frac{u - 1}{u ^{\frac{3}{4}}} : u^{\frac{1}{4}} - \frac{1}{u ^{\frac{3}{4}}}

= 2 \cdot \frac{1}{32} \cdot \int u^{\frac{1}{4}} - \frac{1}{u ^{\frac{3}{4}}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot \frac{1}{32} (\int u^{\frac{1}{4}} d u - \int \frac{1}{u ^{\frac{3}{4}}} d u)

\int u^{\frac{1}{4}} d u = \frac{4}{5} u^{\frac{5}{4}}

\int \frac{1}{u ^{\frac{3}{4}}} d u = 4 u^{\frac{1}{4}}

= 2 \cdot \frac{1}{32} (\frac{4}{5} u^{\frac{5}{4}} - 4 u^{\frac{1}{4}})

Substitute back

u = 1 - 4 x^{2}

= 2 \cdot \frac{1}{32} (\frac{4}{5} (1 - 4 x^{2})^{\frac{5}{4}} - 4 (1 - 4 x^{2})^{\frac{1}{4}})

Simplify

2 \cdot \frac{1}{32} (\frac{4}{5} (1 - 4 x^{2})^{\frac{5}{4}} - 4 (1 - 4 x^{2})^{\frac{1}{4}}) : - \frac{1}{5} (x^{2} + 1) (- 4 x^{2} + 1)^{\frac{1}{4}}

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

Add a constant to the solution

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

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