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

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

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

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

1

Solution steps

\int_{- 2}^{e - 3} \frac{1}{x + 3} 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_{c o s (y)}^{0} e^{x} sin (y) d x

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

\int_{0}^{y} e^{- (x + y)} d x

\int_{0}^{2} x^{2} + 4 x d x

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

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