What is the integral of (3v)/(1-4v^2) ?

The integral of (3v)/(1-4v^2) is -3/8 ln|1-4v^2|+C

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integral of (3v)/(1-4v^2)

\int \frac{3 v}{1 - 4 v ^{2}} d v

- \frac{3}{8} ln 1 - 4 v^{2} + C

Solution steps

\int \frac{3 v}{1 - 4 v ^{2}} d v

Take the constant out:

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

= 3 \cdot \int \frac{v}{1 - 4 v ^{2}} d v

Apply u-substitution

: \int - \frac{1}{8 u} d u

= 3 \cdot \int - \frac{1}{8 u} d u

Take the constant out:

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

= 3 (- \frac{1}{8} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 (- \frac{1}{8} ln ∣ u ∣)

Substitute back

u = 1 - 4 v^{2}

= 3 (- \frac{1}{8} ln 1 - 4 v^{2})

Simplify

3 (- \frac{1}{8} ln 1 - 4 v^{2}) : - \frac{3}{8} ln 1 - 4 v^{2}

= - \frac{3}{8} ln 1 - 4 v^{2}

Add a constant to the solution

= - \frac{3}{8} ln 1 - 4 v^{2} + C

f (x) = x^{5} + 3 x^{2} + 2 x, at x = - 1

3 x^{2} - 4 x + 6

x \to - \infty lim (- 2 e^{x})

\frac{\partial}{\partial y} (\frac{x - y}{x y + 2})

f (5) = 4 x^{2} + 3 x - \frac{8}{x ^{2}}

What is the integral of (3v)/(1-4v^2) ?
The integral of (3v)/(1-4v^2) is -3/8 ln|1-4v^2|+C