What is the integral from-1 to e-2 of 1/(t+2) ?

The integral from-1 to e-2 of 1/(t+2) is 1

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integral from-1 to e-2 of 1/(t+2)

\int_{- 1}^{e - 2} \frac{1}{t + 2} d t

1

Solution steps

\int_{- 1}^{e - 2} \frac{1}{t + 2} d t

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_{2}^{\infty} \frac{2}{1 + x} d x

\int_{0}^{\frac{π}{4}} (3 sec^{2} (x)) d x

\int_{0}^{1} \frac{e ^{z} + 9}{e ^{z} + 9 z} d z

\int_{0}^{4} 2 x - 6 d x

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

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