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

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

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

\int_{0}^{1} (\frac{1}{x + 1}) d x

ln (2)

Decimal

0.69314 \dots

Solution steps

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

Apply u-substitution

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

= \int_{1}^{2} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{2}

Compute the boundaries:

ln (2)

= ln (2)

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

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

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

\int_{1}^{8} 1 + x^{- \frac{2}{3}} d x

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

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