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

The integral of (x^3)/((5-x^4)^6) is 1/(20(5-x^4)^5)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{20 ( 5 - x ^{4} ) ^{5}} + C

Solution steps

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

Apply u-substitution

: \int - \frac{1}{4 u ^{6}} d u

= \int - \frac{1}{4 u ^{6}} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int \frac{1}{u ^{6}} d u

Apply the Power Rule

: - \frac{1}{5 u ^{5}}

= - \frac{1}{4} (- \frac{1}{5 u ^{5}})

Substitute back

u = 5 - x^{4}

= - \frac{1}{4} (- \frac{1}{5 ( 5 - x ^{4} ) ^{5}})

Simplify

- \frac{1}{4} (- \frac{1}{5 ( 5 - x ^{4} ) ^{5}}) : \frac{1}{20 ( 5 - x ^{4} ) ^{5}}

= \frac{1}{20 ( 5 - x ^{4} ) ^{5}}

Add a constant to the solution

= \frac{1}{20 ( 5 - x ^{4} ) ^{5}} + C

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

\int_{0}^{2 π} cos^{2} (6 x) d x

\frac{\partial}{\partial x} (2 x^{3} + 3 y^{2} - 6 x y + 2)

\frac{d}{d x} ((2 x^{4} - x + 1) (- x^{5} + 4))

\int 5 x^{9} d x

What is the integral of (x^3)/((5-x^4)^6) ?
The integral of (x^3)/((5-x^4)^6) is 1/(20(5-x^4)^5)+C