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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{2} ln ∣ u ∣

Substitute back

u = 1 - x^{2}

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

Add a constant to the solution

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

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

\int [5 x^{4} + 10 cos (x)] d x

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\frac{\partial}{\partial x} (3 (x - 6)^{3})

3 - 11 x^{5}

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