What is the integral from 0 to 4 of 5/(3x+1) ?

The integral from 0 to 4 of 5/(3x+1) is 5/3 ln(13)

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integral from 0 to 4 of 5/(3x+1)

\int_{0}^{4} \frac{5}{3 x + 1} d x

\frac{5}{3} ln (13)

Decimal

4.27491 \dots

Solution steps

\int_{0}^{4} \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}^{4} \frac{1}{3 x + 1} d x

Apply u-substitution

= 5 \cdot \int_{1}^{13} \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}^{13} \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}^{13}

Simplify

5 \cdot \frac{1}{3} [ln ∣ u ∣]_{1}^{13} : \frac{5}{3} [ln ∣ u ∣]_{1}^{13}

= \frac{5}{3} [ln ∣ u ∣]_{1}^{13}

Compute the boundaries:

ln (13)

= \frac{5}{3} ln (13)

3^{x}

y^{^{'}} = - \frac{y}{x} + \frac{e ^{x}}{x}

\frac{d y}{d x} = \frac{( x - y )}{x + 2 y}

\int_{0}^{1} 2 (1 + x)^{7} d x

y = - 4 sin (x)

What is the integral from 0 to 4 of 5/(3x+1) ?
The integral from 0 to 4 of 5/(3x+1) is 5/3 ln(13)