What is the integral of (3x^4)/(sqrt(1-6x^5)) ?

The integral of (3x^4)/(sqrt(1-6x^5)) is -1/5 sqrt(1-6x^5)+C

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integral of (3x^4)/(sqrt(1-6x^5))

\int \frac{3 x ^{4}}{1 - 6 x ^{5}} d x

- \frac{1}{5} 1 - 6 x^{5} + C

Solution steps

\int \frac{3 x ^{4}}{1 - 6 x ^{5}} 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 ^{4}}{1 - 6 x ^{5}} d x

Apply u-substitution

: \int - \frac{1}{30 u} d u

= 3 \cdot \int - \frac{1}{30 u} d u

Take the constant out:

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

= 3 (- \frac{1}{30} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

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

= 3 (- \frac{1}{30} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 1 - 6 x^{5}

= 3 (- \frac{1}{30} \cdot 2 (1 - 6 x^{5})^{\frac{1}{2}})

Simplify

3 (- \frac{1}{30} \cdot 2 (1 - 6 x^{5})^{\frac{1}{2}}) : - \frac{1}{5} 1 - 6 x^{5}

= - \frac{1}{5} 1 - 6 x^{5}

Add a constant to the solution

= - \frac{1}{5} 1 - 6 x^{5} + C

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What is the integral of (3x^4)/(sqrt(1-6x^5)) ?
The integral of (3x^4)/(sqrt(1-6x^5)) is -1/5 sqrt(1-6x^5)+C