What is the integral of (2x^5)/(sqrt(1-6x^6)) ?

The integral of (2x^5)/(sqrt(1-6x^6)) is -1/9 sqrt(1-6x^6)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x^5)/(sqrt(1-6x^6))

\int \frac{2 x ^{5}}{1 - 6 x ^{6}} d x

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

Solution steps

\int \frac{2 x ^{5}}{1 - 6 x ^{6}} 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 ^{5}}{1 - 6 x ^{6}} d x

Apply u-substitution

: \int - \frac{1}{36 u} d u

= 2 \cdot \int - \frac{1}{36 u} d u

Take the constant out:

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

= 2 (- \frac{1}{36} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

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

= 2 (- \frac{1}{36} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 1 - 6 x^{6}

= 2 (- \frac{1}{36} \cdot 2 (1 - 6 x^{6})^{\frac{1}{2}})

Simplify

2 (- \frac{1}{36} \cdot 2 (1 - 6 x^{6})^{\frac{1}{2}}) : - \frac{1}{9} 1 - 6 x^{6}

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

Add a constant to the solution

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

f (x) = sin (x)

\frac{d}{d x} (- \frac{x}{8})

\int \frac{x ^{2}}{16 - x ^{2}} d x

\int (\frac{( 3 x + 2 )}{19 - 5 x + x ^{2}}) d x

\int \frac{12 x ^{2}}{2 x ^{3} + 1} d x

What is the integral of (2x^5)/(sqrt(1-6x^6)) ?
The integral of (2x^5)/(sqrt(1-6x^6)) is -1/9 sqrt(1-6x^6)+C