What is the integral from-2 to 1 of 3/(4+x) ?

The integral from-2 to 1 of 3/(4+x) is 3(ln(5)-ln(2))

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integral from-2 to 1 of 3/(4+x)

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

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

Decimal Notation

2.74887 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int_{2}^{5} \frac{1}{u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 3 [ln ∣ u ∣]_{2}^{5}

Compute the boundaries:

ln (5) - ln (2)

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

\int_{2}^{4} sin (x) d x

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\int_{\frac{π}{3}}^{2 π} sin (x) d x

What is the integral from-2 to 1 of 3/(4+x) ?
The integral from-2 to 1 of 3/(4+x) is 3(ln(5)-ln(2))