What is the integral from 0 to s of 1/(0.2s+10) ?

The integral from 0 to s of 1/(0.2s+10) is 5(ln|0.2s+10|-2.30258…)

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integral from 0 to s of 1/(0.2s+10)

\int_{0}^{s} \frac{1}{0.2 s + 10} d s

5 (ln ∣ 0.2 s + 10 ∣ - 2.30258 \dots)

Solution steps

\int_{0}^{s} \frac{1}{0.2 s + 10} d s

Apply u-substitution

= \int_{10}^{0.2 s + 10} \frac{1}{0.2 u} d u

Take the constant out:

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

= \frac{1}{0.2} \cdot \int_{10}^{0.2 s + 10} \frac{1}{u} d u

Simplify

= 5 \cdot \int_{10}^{0.2 s + 10} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 [ln ∣ u ∣]_{10}^{0.2 s + 10}

Compute the boundaries:

ln ∣ 0.2 s + 10 ∣ - 2.30258 \dots

= 5 (ln ∣ 0.2 s + 10 ∣ - 2.30258 \dots)

What is the integral from 0 to s of 1/(0.2s+10) ?
The integral from 0 to s of 1/(0.2s+10) is 5(ln|0.2s+10|-2.30258…)