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

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

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

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

- ln 1 - v^{2} + C

Solution steps

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

Take the constant out:

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

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

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 - v^{2}

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

Simplify

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

= - ln 1 - v^{2}

Add a constant to the solution

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

\int_{0}^{100} e^{- 0.01 x} d x

\frac{x y ^{^{'}}}{e ^{x} + 1} = 2

f (x) = 3 x^{2} - 24 x + 45

x^{2} - 6 x y + y^{2} = 6

f (x) = 2 cos (2 x)

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