What is the integral of (2x^4)/(sqrt(7-6x^5)) ?

The integral of (2x^4)/(sqrt(7-6x^5)) is -2/15 sqrt(7-6x^5)+C

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integral of (2x^4)/(sqrt(7-6x^5))

\int \frac{2 x ^{4}}{7 - 6 x ^{5}} d x

- \frac{2}{15} 7 - 6 x^{5} + C

Solution steps

\int \frac{2 x ^{4}}{7 - 6 x ^{5}} 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 ^{4}}{7 - 6 x ^{5}} d x

Apply u-substitution

: \int - \frac{1}{15} d u

= 2 \cdot \int - \frac{1}{15} d u

Integral of a constant:

\int a d x = a x

= 2 (- \frac{1}{15}) u

Substitute back

u = 7 - 6 x^{5}

= 2 (- \frac{1}{15}) 7 - 6 x^{5}

Simplify

2 (- \frac{1}{15}) 7 - 6 x^{5} : - \frac{2}{15} 7 - 6 x^{5}

= - \frac{2}{15} 7 - 6 x^{5}

Add a constant to the solution

= - \frac{2}{15} 7 - 6 x^{5} + C

\int_{0}^{8 π} 5 + t^{2} d t

\frac{x}{e ^{x}}

f (t) = cos (t)

n = 0 \sum \infty \frac{1 0 ^{n}}{n !}

y^{^{'}} = - 5 y^{2}

What is the integral of (2x^4)/(sqrt(7-6x^5)) ?
The integral of (2x^4)/(sqrt(7-6x^5)) is -2/15 sqrt(7-6x^5)+C