What is the integral from-2 to e-3 of 1/(t+3) ?

The integral from-2 to e-3 of 1/(t+3) is 1

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integral from-2 to e-3 of 1/(t+3)

\int_{- 2}^{e - 3} \frac{1}{t + 3} d t

1

Solution steps

\int_{- 2}^{e - 3} \frac{1}{t + 3} 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_{- 2}^{2} x cos (\frac{nπ x}{2}) d x

\int_{1}^{e} x^{- 1} d x

\int_{0}^{1} x^{2} 4 - x^{2} d x

\int_{0}^{π} 6 e^{x} + 7 sin (x) d x

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

What is the integral from-2 to e-3 of 1/(t+3) ?
The integral from-2 to e-3 of 1/(t+3) is 1