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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \int - \frac{1}{2 u ^{2}} d u

Take the constant out:

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

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

Apply the Power Rule

: - \frac{1}{u}

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

Substitute back

u = 1 - x^{2}

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

Simplify

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

= \frac{1}{1 - x ^{2}}

Add a constant to the solution

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

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