What is the integral of (y^3)/((1-2y^4)^5) ?

The integral of (y^3)/((1-2y^4)^5) is 1/(32(1-2y^4)^4)+C

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integral of (y^3)/((1-2y^4)^5)

\int \frac{y ^{3}}{( 1 - 2 y ^{4} ) ^{5}} d y

\frac{1}{32 ( 1 - 2 y ^{4} ) ^{4}} + C

Solution steps

\int \frac{y ^{3}}{( 1 - 2 y ^{4} ) ^{5}} d y

Apply u-substitution

: \int - \frac{1}{8 u ^{5}} d u

= \int - \frac{1}{8 u ^{5}} d u

Take the constant out:

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

= - \frac{1}{8} \cdot \int \frac{1}{u ^{5}} d u

Apply the Power Rule

: - \frac{1}{4 u ^{4}}

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

Substitute back

u = 1 - 2 y^{4}

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

Simplify

- \frac{1}{8} (- \frac{1}{4 ( 1 - 2 y ^{4} ) ^{4}}) : \frac{1}{32 ( 1 - 2 y ^{4} ) ^{4}}

= \frac{1}{32 ( 1 - 2 y ^{4} ) ^{4}}

Add a constant to the solution

= \frac{1}{32 ( 1 - 2 y ^{4} ) ^{4}} + C

\int_{0}^{1} \frac{36}{x ^{2} - 36} d x

x \to 3 lim (\frac{x - 3}{x + 4})

n = 1 \sum \infty \frac{3 n}{5 n - 1}

x^{\frac{1}{2}}, 4

\frac{d}{d x} (2 x + \frac{π}{2})

What is the integral of (y^3)/((1-2y^4)^5) ?
The integral of (y^3)/((1-2y^4)^5) is 1/(32(1-2y^4)^4)+C