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))

integral

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

- 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

ln (3 x - 2)

\frac{( 1 + 2 x )}{( 1 + x )}

\frac{1}{( x ^{2} - 9 ) ^{2}}

\frac{9}{x ^{5}}

3 x + 2

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