What is the integral from 0 to 3 of 1/(3x+1) ?

The integral from 0 to 3 of 1/(3x+1) is 1/3 ln(10)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 3 of 1/(3x+1)

\int_{0}^{3} \frac{1}{3 x + 1} d x

\frac{1}{3} ln (10)

Decimal

0.76752 \dots

Solution steps

\int_{0}^{3} \frac{1}{3 x + 1} d x

Apply u-substitution

: \int_{1}^{10} \frac{1}{3 u} d u

= \int_{1}^{10} \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_{1}^{10} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{1}^{10}

Compute the boundaries:

ln (10)

= \frac{1}{3} ln (10)

\int \frac{1}{1 + 5 x ^{2}} d x

3 - 5 x

x^{2} + 5 x + 1

\int cot (11 t) 1 - sin^{2} (11 t) d t

f (x) = a x^{2} + b x

What is the integral from 0 to 3 of 1/(3x+1) ?
The integral from 0 to 3 of 1/(3x+1) is 1/3 ln(10)