What is the integral of (x^3)/((1-x^4)^6) ?

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

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integral of (x^3)/((1-x^4)^6)

\int \frac{x ^{3}}{( 1 - x ^{4} ) ^{6}} d x

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

Solution steps

\int \frac{x ^{3}}{( 1 - 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 = 1 - x^{4}

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (x^{2} e^{x} - 2 x e^{x} + 8 e^{x})

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

x \to 0 lim (\frac{6 ^{x} - 1}{x})

\frac{d}{d x} (e^{x - 7})

y^{^{''}} - y^{^{'}} = 2 x

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