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

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

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

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

- ln 1 - u^{2} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \int - \frac{1}{2 v} d v

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}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 (- \frac{1}{2} ln ∣ v ∣)

Substitute back

v = 1 - u^{2}

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

Simplify

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

= - ln 1 - u^{2}

Add a constant to the solution

= - ln 1 - u^{2} + C

(- 9, 5), (- 3, 3)

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