What is the integral of 1/(200+3t) ?

The integral of 1/(200+3t) is 1/3 ln|200+3t|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(200+3t)

\int \frac{1}{200 + 3 t} d t

\frac{1}{3} ln ∣ 200 + 3 t ∣ + C

Solution steps

\int \frac{1}{200 + 3 t} d t

Apply u-substitution

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

= \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 = 200 + 3 t

= \frac{1}{3} ln ∣ 200 + 3 t ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 200 + 3 t ∣ + C

n = 3 \sum \infty \frac{4}{n ^{2} - 1}

x \to 0 lim (- 3 e^{3 x})

\int x e^{- \frac{x}{4}} d x

x + \frac{1}{x}, at x = 4

\int ln^{2} (c) os x tan (x)

What is the integral of 1/(200+3t) ?
The integral of 1/(200+3t) is 1/3 ln|200+3t|+C