What is the integral from 0 to 3 of 5/(3x+1) ?

The integral from 0 to 3 of 5/(3x+1) is 5/3 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 5/(3x+1)

\int_{0}^{3} \frac{5}{3 x + 1} d x

\frac{5}{3} ln (10)

Decimal

3.83764 \dots

Solution steps

\int_{0}^{3} \frac{5}{3 x + 1} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{3} \frac{1}{3 x + 1} d x

Apply u-substitution

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

= 5 \cdot \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

= 5 \cdot \frac{1}{3} \cdot \int_{1}^{10} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{3} [ln ∣ u ∣]_{1}^{10}

Simplify

5 \cdot \frac{1}{3} [ln ∣ u ∣]_{1}^{10} : \frac{5}{3} [ln ∣ u ∣]_{1}^{10}

= \frac{5}{3} [ln ∣ u ∣]_{1}^{10}

Compute the boundaries:

ln (10)

= \frac{5}{3} ln (10)

\int_{1}^{4} 63 x^{\frac{5}{2}} d x

\int_{2}^{2} g (x) d x

\int_{1}^{3} (\frac{x ^{2} + 4}{x}) d x

\int_{1}^{3} (\frac{x ^{2} - 1}{4 x ^{2}}) d x

\int_{0}^{\frac{π}{3}} sin (θ) d θ

What is the integral from 0 to 3 of 5/(3x+1) ?
The integral from 0 to 3 of 5/(3x+1) is 5/3 ln(10)