What is the integral of (x^3)/((1-x^4)^7) ?

The integral of (x^3)/((1-x^4)^7) is 1/(24(1-x^4)^6)+C

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integral of (x^3)/((1-x^4)^7)

\int \frac{x ^{3}}{( 1 - x ^{4} ) ^{7}} d x

\frac{1}{24 ( 1 - x ^{4} ) ^{6}} + C

Solution steps

\int \frac{x ^{3}}{( 1 - x ^{4} ) ^{7}} d x

Apply u-substitution

: \int - \frac{1}{4 u ^{7}} d u

= \int - \frac{1}{4 u ^{7}} 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 ^{7}} d u

Apply the Power Rule

: - \frac{1}{6 u ^{6}}

= - \frac{1}{4} (- \frac{1}{6 u ^{6}})

Substitute back

u = 1 - x^{4}

= - \frac{1}{4} (- \frac{1}{6 ( 1 - x ^{4} ) ^{6}})

Simplify

- \frac{1}{4} (- \frac{1}{6 ( 1 - x ^{4} ) ^{6}}) : \frac{1}{24 ( 1 - x ^{4} ) ^{6}}

= \frac{1}{24 ( 1 - x ^{4} ) ^{6}}

Add a constant to the solution

= \frac{1}{24 ( 1 - x ^{4} ) ^{6}} + C

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What is the integral of (x^3)/((1-x^4)^7) ?
The integral of (x^3)/((1-x^4)^7) is 1/(24(1-x^4)^6)+C