What is the integral of (4x^3)/((1-2x^4)^6) ?

The integral of (4x^3)/((1-2x^4)^6) is 1/(10(-2x^4+1)^5)+C

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integral of (4x^3)/((1-2x^4)^6)

\int \frac{4 x ^{3}}{( 1 - 2 x ^{4} ) ^{6}} d x

\frac{1}{10 ( - 2 x ^{4} + 1 ) ^{5}} + C

Solution steps

\int \frac{4 x ^{3}}{( 1 - 2 x ^{4} ) ^{6}} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{3}}{( 1 - 2 x ^{4} ) ^{6}} d x

Apply u-substitution

: \int - \frac{1}{8 u ^{6}} d u

= 4 \cdot \int - \frac{1}{8 u ^{6}} d u

Take the constant out:

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

= 4 (- \frac{1}{8} \cdot \int \frac{1}{u ^{6}} d u)

Apply the Power Rule

: - \frac{1}{5 u ^{5}}

= 4 (- \frac{1}{8} (- \frac{1}{5 u ^{5}}))

Substitute back

u = 1 - 2 x^{4}

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

Simplify

4 (- \frac{1}{8} (- \frac{1}{5 ( 1 - 2 x ^{4} ) ^{5}})) : \frac{1}{10 ( - 2 x ^{4} + 1 ) ^{5}}

= \frac{1}{10 ( - 2 x ^{4} + 1 ) ^{5}}

Add a constant to the solution

= \frac{1}{10 ( - 2 x ^{4} + 1 ) ^{5}} + C

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What is the integral of (4x^3)/((1-2x^4)^6) ?
The integral of (4x^3)/((1-2x^4)^6) is 1/(10(-2x^4+1)^5)+C