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

The integral from 0 to t of 1/(x+10) is ln|10+t|-ln(10)

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

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

ln ∣ 10 + t ∣ - ln (10)

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{10}^{10 + t}

Compute the boundaries:

ln ∣ 10 + t ∣ - ln (10)

= ln ∣ 10 + t ∣ - ln (10)

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