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

Take the constant out:

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

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

Apply u-substitution

= - \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})

Simplify

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

Add a constant to the solution

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

\int t t - 1 d t

(cosh (x))^{^{'}}

\frac{d}{d x} (25)

x \to 0 - lim (x e^{\frac{2}{x}})

\int_{0}^{2} x e^{- x^{2}} d 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