What is the integral of-(8x^3)/((5-2x^4)^4) ?

The integral of-(8x^3)/((5-2x^4)^4) is -1/(3(5-2x^4)^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of-(8x^3)/((5-2x^4)^4)

\int - \frac{8 x ^{3}}{( 5 - 2 x ^{4} ) ^{4}} d x

- \frac{1}{3 ( 5 - 2 x ^{4} ) ^{3}} + C

Solution steps

\int - \frac{8 x ^{3}}{( 5 - 2 x ^{4} ) ^{4}} d x

Take the constant out:

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

= - 8 \cdot \int \frac{x ^{3}}{( 5 - 2 x ^{4} ) ^{4}} d x

Apply u-substitution

= - 8 \cdot \int - \frac{1}{8 u ^{4}} d u

Take the constant out:

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

= - 8 (- \frac{1}{8} \cdot \int \frac{1}{u ^{4}} d u)

Apply the Power Rule

= - 8 (- \frac{1}{8} (- \frac{1}{3 u ^{3}}))

Substitute back

u = 5 - 2 x^{4}

= - 8 (- \frac{1}{8} (- \frac{1}{3 ( 5 - 2 x ^{4} ) ^{3}}))

Simplify

- 8 (- \frac{1}{8} (- \frac{1}{3 ( 5 - 2 x ^{4} ) ^{3}})) : - \frac{1}{3 ( 5 - 2 x ^{4} ) ^{3}}

= - \frac{1}{3 ( 5 - 2 x ^{4} ) ^{3}}

Add a constant to the solution

= - \frac{1}{3 ( 5 - 2 x ^{4} ) ^{3}} + C

What is the integral of-(8x^3)/((5-2x^4)^4) ?
The integral of-(8x^3)/((5-2x^4)^4) is -1/(3(5-2x^4)^3)+C