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

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

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

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

1

Solution steps

\int_{- 3}^{e - 4} \frac{1}{t + 4} 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_{0}^{2} 5 + x d x

\int_{x}^{y} x d x

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

\int_{0}^{2} (6 x - 3) (8 x^{2} + 5) d x

\int_{1}^{5} \frac{1}{( x + 2 ) ( x + 3 )} d x

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