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

The integral from 0 to 3 of 1/(5x+1) is 4/5 ln(2)

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

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

\frac{4}{5} ln (2)

Decimal Notation

0.55451 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{16} \frac{1}{5 u} d u

= \int_{1}^{16} \frac{1}{5 u} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int_{1}^{16} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{5} [ln ∣ u ∣]_{1}^{16}

Compute the boundaries:

4 ln (2)

= \frac{1}{5} \cdot 4 ln (2)

Simplify

= \frac{4}{5} ln (2)

\int_{0}^{5} (5 x - 1) d x

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

y = ∣ x ∣

7 x + 1, at x = 9

cos (t)

What is the integral from 0 to 3 of 1/(5x+1) ?
The integral from 0 to 3 of 1/(5x+1) is 4/5 ln(2)