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

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

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

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

\frac{1}{16 ( 5 - x ^{4} ) ^{4}} + C

Solution steps

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

Apply u-substitution

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

= \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 u ^{4}}

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

Substitute back

u = 5 - x^{4}

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

Simplify

- \frac{1}{4} (- \frac{1}{4 ( 5 - x ^{4} ) ^{4}}) : \frac{1}{16 ( 5 - x ^{4} ) ^{4}}

= \frac{1}{16 ( 5 - x ^{4} ) ^{4}}

Add a constant to the solution

= \frac{1}{16 ( 5 - x ^{4} ) ^{4}} + C

y = 4 (x - \frac{1}{x})^{3}, at x = 2

\int \frac{1}{( x - 1 ) 4 x ^{2} - 8 x + 3} d x

\int x + \frac{2}{sin ^{2} ( x )} d x

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

\frac{d}{d x} (8 cos^{4} (x))

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