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

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

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

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

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

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

Apply u-substitution

: \int_{3}^{6} \frac{1}{u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{3}^{6}

Compute the boundaries:

ln (6) - ln (3)

= ln (6) - ln (3)

Simplify

= ln (2)

\int_{0}^{\frac{π}{2}} (sin (x))^{6} d x

\int_{0}^{1} y^{3} - y^{2} d y

\int_{- 1}^{3} x^{- 3} d x

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

\int_{0}^{- \infty} e^{3 x} d x

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