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

The integral of x/((4-2x^2)^3) is 1/(8(4-2x^2)^2)+C

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

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{1}{4 u ^{3}} d u

= \int - \frac{1}{4 u ^{3}} 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 ^{3}} d u

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

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

Substitute back

u = 4 - 2 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{x + 9}{x ^{2} + 4 x + 8} d x

\frac{d}{d x} (\frac{1}{3} x^{3} + \frac{1}{2} x^{2} - 6 x + 8)

\int x^{2} x^{3} + 35 d x

\frac{d}{d x} (1 + 1 - x)

y^{^{'}} - 5 y = 2 e^{6 t}

What is the integral of x/((4-2x^2)^3) ?
The integral of x/((4-2x^2)^3) is 1/(8(4-2x^2)^2)+C