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

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

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

= - 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

in v erse l a pl a ce \frac{2}{s ^{2} + 4}

\int \frac{5 - 3 sin ( 2 x )}{sec ( 6 x )} d x

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

d er i v a t i v e \frac{1}{4 + x ^{2}}

d er i v a t i v e ln (\frac{8 - x}{8 + x})

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