What is the integral from 0 to 1 of 2/(3+x) ?

The integral from 0 to 1 of 2/(3+x) is 2(2ln(2)-ln(3))

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integral from 0 to 1 of 2/(3+x)

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

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

Decimal

0.57536 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{4} \frac{1}{u} d u

= 2 \cdot \int_{3}^{4} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 [ln ∣ u ∣]_{3}^{4}

Compute the boundaries:

2 ln (2) - ln (3)

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

\int_{8}^{27} (3 x - \frac{1}{3 x}) d x

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

\int_{1}^{3} 2 ln (x) d x

\int_{1}^{2} \frac{4}{2} d x

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

What is the integral from 0 to 1 of 2/(3+x) ?
The integral from 0 to 1 of 2/(3+x) is 2(2ln(2)-ln(3))