What is the integral from-3 to e-4 of 1/(x+4) ?

The integral from-3 to e-4 of 1/(x+4) is 1

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integral from-3 to e-4 of 1/(x+4)

\int_{- 3}^{e - 4} \frac{1}{x + 4} d x

1

Solution steps

\int_{- 3}^{e - 4} \frac{1}{x + 4} 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_{\frac{1}{2}}^{2} (3 - \frac{3}{x ^{2}}) d x

\int_{0}^{a} x e^{x^{2}} d x

\int_{0}^{1} \frac{1}{4 x + 3} d x

\int_{0}^{4} ∣ 3 x - 8 ∣ d x

\int_{1}^{x^{2}} sin (x) d x

What is the integral from-3 to e-4 of 1/(x+4) ?
The integral from-3 to e-4 of 1/(x+4) is 1