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

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

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

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

ln 1 - x^{2} + C

Solution steps

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

Apply u-substitution

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

= - 2 \cdot \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 2 (- \frac{1}{2} ln ∣ u ∣)

Substitute back

u = 1 - x^{2}

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

Simplify

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

= ln 1 - x^{2}

Add a constant to the solution

= ln 1 - x^{2} + C

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

\int 3 z d z

\int \frac{7}{x - 3} d x

y^{^{'}} = \frac{2 x sec ( \frac{y}{x} ) + y}{x}

\frac{d}{d x} (\frac{x - 4}{3})

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