What is the integral from-6 to-10 of 1/(x+2) ?

The integral from-6 to-10 of 1/(x+2) is ln(2)

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integral from-6 to-10 of 1/(x+2)

\int_{- 6}^{- 10} \frac{1}{x + 2} d x

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

\int_{- 6}^{- 10} \frac{1}{x + 2} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 10}^{- 6} \frac{1}{x + 2} d x

Apply u-substitution

: \int_{- 8}^{- 4} \frac{1}{u} d u

= - \int_{- 8}^{- 4} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - [ln ∣ u ∣]_{- 8}^{- 4}

Compute the boundaries:

- ln (2)

= - (- ln (2))

Simplify

= ln (2)

f (x) = x^{2} - 5, (3, 4)

y = \frac{1}{x ^{7}}

f (x) = x^{2} - \frac{1}{x}, (- 1, 2)

h \to 0 lim (\frac{( h + 2 ) ^{2} - 4}{h})

y = \frac{2 x}{x + 5}, (5, 1)

What is the integral from-6 to-10 of 1/(x+2) ?
The integral from-6 to-10 of 1/(x+2) is ln(2)