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

The integral from 0 to e of 1/(x+e) is ln(2e)-1

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integral from 0 to e of 1/(x+e)

\int_{0}^{e} \frac{1}{x + e} d x

ln (2 e) - 1

Decimal Notation

0.69314 \dots

Solution steps

\int_{0}^{e} \frac{1}{x + e} d x

Apply u-substitution

: \int_{e}^{2 e} \frac{1}{u} d u

= \int_{e}^{2 e} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{e}^{2 e}

Compute the boundaries:

ln (2 e) - 1

= ln (2 e) - 1

\int \frac{x}{1 + 9 x ^{2}} d x

\frac{e ^{- 2 s}}{s - 9}

\int cos^{3} (3 x) sin^{3} (3 x) d x

f (x) = (e^{x} + x^{π}) (x^{2} + x + 1)

y^{^{'}} = 0.1 y (10 - y)

What is the integral from 0 to e of 1/(x+e) ?
The integral from 0 to e of 1/(x+e) is ln(2e)-1