What is the integral of (4x^5)/(sqrt(1-12x^6)) ?

The integral of (4x^5)/(sqrt(1-12x^6)) is -1/9 sqrt(1-12x^6)+C

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integral of (4x^5)/(sqrt(1-12x^6))

\int \frac{4 x ^{5}}{1 - 12 x ^{6}} d x

- \frac{1}{9} 1 - 12 x^{6} + C

Solution steps

\int \frac{4 x ^{5}}{1 - 12 x ^{6}} 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 ^{5}}{1 - 12 x ^{6}} d x

Apply u-substitution

: \int - \frac{1}{72 u} d u

= 4 \cdot \int - \frac{1}{72 u} d u

Take the constant out:

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

= 4 (- \frac{1}{72} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

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

= 4 (- \frac{1}{72} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 1 - 12 x^{6}

= 4 (- \frac{1}{72} \cdot 2 (1 - 12 x^{6})^{\frac{1}{2}})

Simplify

4 (- \frac{1}{72} \cdot 2 (1 - 12 x^{6})^{\frac{1}{2}}) : - \frac{1}{9} 1 - 12 x^{6}

= - \frac{1}{9} 1 - 12 x^{6}

Add a constant to the solution

= - \frac{1}{9} 1 - 12 x^{6} + C

12 x^{3} - 18 x^{2}

\frac{d y}{d x} + 2 y = k

f (x) = e^{x^{2}}, at x = \frac{1}{2}

\int \frac{1}{p} d p

What is the integral of (4x^5)/(sqrt(1-12x^6)) ?
The integral of (4x^5)/(sqrt(1-12x^6)) is -1/9 sqrt(1-12x^6)+C