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 ^{2}}{d x ^{2}} (\frac{x}{x ^{2} - 6})

\frac{d y}{d x} + 3 y = e^{4 x}

n = 0 \sum \infty (\frac{n + 1}{n})^{n}

3 x^{^{''}} - 6 x^{^{'}} + 3 x = 0

x \to 0 + lim (\frac{5}{x} - \frac{5}{∣ x ∣})

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