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

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

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

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

1

Solution steps

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

Apply u-substitution

= \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}^{π} 5 x sin (x) d x

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

\int_{0}^{a} 5 e^{x} d x

\int_{1}^{2} \frac{1}{x ^{2} + 1} d x

\int_{1}^{2} 3 e^{x} d x

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