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

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

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

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

5 (ln ∣ 0.2 x + 10 ∣ - 2.30258 \dots)

Solution steps

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

Apply u-substitution

: \int_{10}^{0.2 x + 10} \frac{1}{0.2 u} d u

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

Simplify

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 [ln ∣ u ∣]_{10}^{0.2 x + 10}

Compute the boundaries:

ln ∣ 0.2 x + 10 ∣ - 2.30258 \dots

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

1 - x^{2}

\int \frac{3 x + 1}{4 - x ^{2}} d x

\int_{0}^{6} \frac{6 x}{1 + x ^{2}} d x

(x + 5) (7 x + 3)

y^{^{'}} - y = t

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