What is the integral from 0 to 2 of 2/(3+x) ?

The integral from 0 to 2 of 2/(3+x) is 2(ln(5)-ln(3))

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integral from 0 to 2 of 2/(3+x)

\int_{0}^{2} \frac{2}{3 + x} d x

2 (ln (5) - ln (3))

Decimal

1.02165 \dots

Solution steps

\int_{0}^{2} \frac{2}{3 + x} d x

Take the constant out:

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

= 2 \cdot \int_{0}^{2} \frac{1}{3 + x} d x

Apply u-substitution

: \int_{3}^{5} \frac{1}{u} d u

= 2 \cdot \int_{3}^{5} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 [ln ∣ u ∣]_{3}^{5}

Compute the boundaries:

ln (5) - ln (3)

= 2 (ln (5) - ln (3))

\int_{0}^{3} x^{2} sin (\frac{nπ x}{3}) d x

\int_{0}^{2} (2 x^{2} - 3) d x

\int_{0}^{x^{2}} \frac{1}{x + y} d y

\int_{2}^{6} 8 x d x

\int_{0}^{1} 2 x \cdot e^{x^{2}} d x

What is the integral from 0 to 2 of 2/(3+x) ?
The integral from 0 to 2 of 2/(3+x) is 2(ln(5)-ln(3))