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

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

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

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

1

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{e}

Compute the boundaries:

1

= 1

\int_{0}^{0.1} \frac{1}{1 + x ^{4}} d x

\int_{0}^{1} (2 + e^{x})^{2} d x

\int_{- 3}^{1} (2 x + 1) (3 + x) (1 - x) d x

\int_{- 1}^{2} - x^{2} + 4 d x

\int_{1}^{3} (5 x - x^{2}) d x

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