What is the integral of v/(3-0.001v^2) ?

The integral of v/(3-0.001v^2) is -500ln|3-0.001v^2|+C

integral of v/(3-0.001v^2)

\int \frac{v}{3 - 0.001 v ^{2}} d v

- 500 ln 3 - 0.001 v^{2} + C

Solution steps

\int \frac{v}{3 - 0.001 v ^{2}} d v

Apply u-substitution

= \int - \frac{1}{0.002 u} d u

Take the constant out:

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

= - \frac{1}{0.002} \cdot \int \frac{1}{u} d u

Simplify

= - 500 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 500 ln ∣ u ∣

Substitute back

u = 3 - 0.001 v^{2}

= - 500 ln 3 - 0.001 v^{2}

Add a constant to the solution

= - 500 ln 3 - 0.001 v^{2} + C

\frac{\partial}{\partial x} (y + arctan (\frac{y}{x}))

\frac{\partial}{\partial y} (x^{3} - y^{2})

t a y l or ln (6 x + 1)

x \to \infty lim (\frac{e ^{5 x - 3}}{ln ( x - 2 )})

d er i v a t i v e f (x) = cos (a^{8} + x^{8})

What is the integral of v/(3-0.001v^2) ?
The integral of v/(3-0.001v^2) is -500ln|3-0.001v^2|+C