What is the integral from 0 to 3 of (12)/(3x+1) ?

The integral from 0 to 3 of (12)/(3x+1) is 4ln(10)

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integral from 0 to 3 of (12)/(3x+1)

\int_{0}^{3} \frac{12}{3 x + 1} d x

4 ln (10)

Decimal

9.21034 \dots

Solution steps

\int_{0}^{3} \frac{12}{3 x + 1} d x

Take the constant out:

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

= 12 \cdot \int_{0}^{3} \frac{1}{3 x + 1} d x

Apply u-substitution

= 12 \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

= 12 \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 ∣)

= 12 \cdot \frac{1}{3} [ln ∣ u ∣]_{1}^{10}

Simplify

12 \cdot \frac{1}{3} [ln ∣ u ∣]_{1}^{10} : 4 [ln ∣ u ∣]_{1}^{10}

= 4 [ln ∣ u ∣]_{1}^{10}

Compute the boundaries:

ln (10)

= 4 ln (10)

What is the integral from 0 to 3 of (12)/(3x+1) ?
The integral from 0 to 3 of (12)/(3x+1) is 4ln(10)