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

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

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

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

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

Solution steps

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

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 x^{4}

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

Simplify

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

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

Add a constant to the solution

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

x y^{^{''}} = y^{^{'}} + (y^{^{'}})^{3}

x \to - 3 - lim (\frac{4 x}{9 - x ^{2}})

\int \frac{1}{x ^{2} - x + 2} d x

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

(t^{2} + 16)^{2} y^{^{'}} + 2 t (t^{2} + 16) y = 4, y (- 2) = 10

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