What is the integral of v= 1/(3t+2) ?

The integral of v= 1/(3t+2) is 1/3 ln|3t+2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of v= 1/(3t+2)

integral

v = \frac{1}{3 t + 2}

\frac{1}{3} ln ∣ 3 t + 2 ∣ + C

Solution steps

\int \frac{1}{3 t + 2} d t

Apply u-substitution

= \int \frac{1}{3 u} d u

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 3 t + 2

= \frac{1}{3} ln ∣ 3 t + 2 ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 3 t + 2 ∣ + C

in t e g r a l \frac{( x + 1 )}{( x ^{2} + x + 1 )}

in t e g r a l ln (\frac{e ^{x}}{x ^{2}})

in t e g r a l (x^{2} - a^{2})^{\frac{1}{2}}

in t e g r a l 6 e^{- 2} x

in t e g r a l \frac{x ^{3} + 4 x ^{2} - 3 x + 2}{x ^{2}}

What is the integral of v= 1/(3t+2) ?
The integral of v= 1/(3t+2) is 1/3 ln|3t+2|+C