What is the integral of v/(4+v^2) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of v/(4+v^2)

\int \frac{v}{4 + v ^{2}} d v

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

Solution steps

\int \frac{v}{4 + v ^{2}} d v

Apply u-substitution

: \int \frac{1}{2 u} d u

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 4 + v^{2}

= \frac{1}{2} ln 4 + v^{2}

Add a constant to the solution

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

\int x^{4} (x^{5} - 1)^{7} d x

\int \frac{12 x ^{3}}{x ^{4} + 9} d x

\int \frac{2}{x ^{2} + 2 x + 4} d x

\int \frac{1}{( x - 4 ) ( x + 2 )} d x

\int \frac{2}{x ^{2} - 2} d x

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