What is the integral of (x^5)/((8-x^6)^2) ?

The integral of (x^5)/((8-x^6)^2) is 1/(6(8-x^6))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^5)/((8-x^6)^2)

\int \frac{x ^{5}}{( 8 - x ^{6} ) ^{2}} d x

\frac{1}{6 ( 8 - x ^{6} )} + C

Solution steps

\int \frac{x ^{5}}{( 8 - x ^{6} ) ^{2}} d x

Apply u-substitution

: \int - \frac{1}{6 u ^{2}} d u

= \int - \frac{1}{6 u ^{2}} d u

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{u}

= - \frac{1}{6} (- \frac{1}{u})

Substitute back

u = 8 - x^{6}

= - \frac{1}{6} (- \frac{1}{8 - x ^{6}})

Simplify

- \frac{1}{6} (- \frac{1}{8 - x ^{6}}) : \frac{1}{6 ( 8 - x ^{6} )}

= \frac{1}{6 ( 8 - x ^{6} )}

Add a constant to the solution

= \frac{1}{6 ( 8 - x ^{6} )} + C

\int \frac{2}{5 x - 8} d x

\int \frac{( x ^{3} - 4 x - 10 )}{( x ^{2} - x - 6 )} d x

\int \frac{5}{( t ^{2} + 1 ) ^{\frac{3}{2}}} d t

\frac{d}{d x} (- \frac{7}{x ^{4}})

y = \frac{x}{x - 1}

What is the integral of (x^5)/((8-x^6)^2) ?
The integral of (x^5)/((8-x^6)^2) is 1/(6(8-x^6))+C