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

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

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

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

1

Solution steps

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

\int_{0}^{7} (2 e^{x}) d x

\int_{0}^{25} 85 - 3 25 - t d t

\int_{0}^{x} \frac{t ^{2}}{t ^{4} + 1} d t

\int_{0}^{5} \frac{1}{4} x d x

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