What is the integral of (3x^2)/(sqrt(1-6x^3)) ?

The integral of (3x^2)/(sqrt(1-6x^3)) is -1/3 sqrt(1-6x^3)+C

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integral of (3x^2)/(sqrt(1-6x^3))

\int \frac{3 x ^{2}}{1 - 6 x ^{3}} d x

- \frac{1}{3} 1 - 6 x^{3} + C

Solution steps

\int \frac{3 x ^{2}}{1 - 6 x ^{3}} d x

Take the constant out:

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

= 3 \cdot \int \frac{x ^{2}}{1 - 6 x ^{3}} d x

Apply u-substitution

: \int - \frac{1}{18 u} d u

= 3 \cdot \int - \frac{1}{18 u} d u

Take the constant out:

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

= 3 (- \frac{1}{18} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

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

= 3 (- \frac{1}{18} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 1 - 6 x^{3}

= 3 (- \frac{1}{18} \cdot 2 (1 - 6 x^{3})^{\frac{1}{2}})

Simplify

3 (- \frac{1}{18} \cdot 2 (1 - 6 x^{3})^{\frac{1}{2}}) : - \frac{1}{3} 1 - 6 x^{3}

= - \frac{1}{3} 1 - 6 x^{3}

Add a constant to the solution

= - \frac{1}{3} 1 - 6 x^{3} + C

\frac{\partial}{\partial x} (y sin (z))

x \to - 8 - lim (\frac{x + 2}{x + 8})

8 x - 10, x^{2} - 4 x + 10

\frac{d y}{d x}, e^{x} y = 1

\int x^{2} - 2 d x

What is the integral of (3x^2)/(sqrt(1-6x^3)) ?
The integral of (3x^2)/(sqrt(1-6x^3)) is -1/3 sqrt(1-6x^3)+C