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} (sec (x) - 1)

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

f (x) = cos (x) e^{x}

\int \frac{x}{1 + x ^{3}} d x

\frac{d}{d x} (3 + 3 cos (x^{3}) - 4 sin (x^{3}))

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