What is the integral of (x^3)/((7-x^4)^5) ?

The integral of (x^3)/((7-x^4)^5) is 1/(16(7-x^4)^4)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (x^3)/((7-x^4)^5)

\int \frac{x ^{3}}{( 7 - x ^{4} ) ^{5}} d x

\frac{1}{16 ( 7 - x ^{4} ) ^{4}} + C

Solution steps

\int \frac{x ^{3}}{( 7 - x ^{4} ) ^{5}} d x

Apply u-substitution

= \int - \frac{1}{4 u ^{5}} 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 ^{5}} d u

Apply the Power Rule

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

Substitute back

u = 7 - x^{4}

= - \frac{1}{4} (- \frac{1}{4 ( 7 - x ^{4} ) ^{4}})

Simplify

- \frac{1}{4} (- \frac{1}{4 ( 7 - x ^{4} ) ^{4}}) : \frac{1}{16 ( 7 - x ^{4} ) ^{4}}

= \frac{1}{16 ( 7 - x ^{4} ) ^{4}}

Add a constant to the solution

= \frac{1}{16 ( 7 - x ^{4} ) ^{4}} + C

What is the integral of (x^3)/((7-x^4)^5) ?
The integral of (x^3)/((7-x^4)^5) is 1/(16(7-x^4)^4)+C