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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 ( 1 - 2 u ) ^{3}} d u

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

Take the constant out:

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

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

Apply u-substitution

: \int - \frac{1}{2 v ^{3}} d v

= - 4 \cdot \frac{1}{2} \cdot \int - \frac{1}{2 v ^{3}} d v

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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