What is the integral of (4x^2)/(sqrt(1-8x^3)) ?

The integral of (4x^2)/(sqrt(1-8x^3)) is -1/3 sqrt(1-8x^3)+C

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integral of (4x^2)/(sqrt(1-8x^3))

\int \frac{4 x ^{2}}{1 - 8 x ^{3}} d x

- \frac{1}{3} 1 - 8 x^{3} + C

Solution steps

\int \frac{4 x ^{2}}{1 - 8 x ^{3}} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{2}}{1 - 8 x ^{3}} d x

Apply u-substitution

: \int - \frac{1}{24 u} d u

= 4 \cdot \int - \frac{1}{24 u} d u

Take the constant out:

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

= 4 (- \frac{1}{24} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

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

= 4 (- \frac{1}{24} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 1 - 8 x^{3}

= 4 (- \frac{1}{24} \cdot 2 (1 - 8 x^{3})^{\frac{1}{2}})

Simplify

4 (- \frac{1}{24} \cdot 2 (1 - 8 x^{3})^{\frac{1}{2}}) : - \frac{1}{3} 1 - 8 x^{3}

= - \frac{1}{3} 1 - 8 x^{3}

Add a constant to the solution

= - \frac{1}{3} 1 - 8 x^{3} + C

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What is the integral of (4x^2)/(sqrt(1-8x^3)) ?
The integral of (4x^2)/(sqrt(1-8x^3)) is -1/3 sqrt(1-8x^3)+C