What is the integral of 1/(300+x) ?

The integral of 1/(300+x) is ln|300+x|+C

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integral of 1/(300+x)

\int \frac{1}{300 + x} d x

ln ∣ 300 + x ∣ + C

Solution steps

\int \frac{1}{300 + x} d x

Apply u-substitution

: \int \frac{1}{u} d u

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = 300 + x

= ln ∣ 300 + x ∣

Add a constant to the solution

= ln ∣ 300 + x ∣ + C

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