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

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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

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

= - 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

: - \frac{1}{3 u ^{3}}

= - 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

(3 x + 2)^{^{'}}

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

10 y^{^{''}} - y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 0

\frac{d}{d x} (\frac{1 - 0.6 x}{x})

\frac{\partial}{\partial x} (\frac{x}{1 + x y})

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