What is the integral from 4 to 6 of-3/(x+2) ?

The integral from 4 to 6 of-3/(x+2) is -3ln(4/3)

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integral from 4 to 6 of-3/(x+2)

\int_{4}^{6} - \frac{3}{x + 2} d x

- 3 ln (\frac{4}{3})

Solution steps

\int_{4}^{6} - \frac{3}{x + 2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - 3 \cdot \int_{4}^{6} \frac{1}{x + 2} d x

Apply u-substitution

: \int_{6}^{8} \frac{1}{u} d u

= - 3 \cdot \int_{6}^{8} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 3 [ln ∣ u ∣]_{6}^{8}

Compute the boundaries:

3 ln (2) - ln (6)

= - 3 (3 ln (2) - ln (6))

Simplify

= - 3 ln (\frac{4}{3})

\int_{1}^{2} \frac{e ^{2 l n (x)}}{x} d x

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

\int_{1}^{8} x^{2} - 4 x - 12 d x

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

\int_{0}^{5} (x^{2} - 4) d x

What is the integral from 4 to 6 of-3/(x+2) ?
The integral from 4 to 6 of-3/(x+2) is -3ln(4/3)