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

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

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

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

ln (3)

Decimal

1.09861 \dots

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= [ln ∣ u ∣]_{1}^{3}

Compute the boundaries:

ln (3)

= ln (3)

\int_{- 1}^{1} e^{u + 1} d u

x \to 2 lim (\frac{x ^{2} + 1}{x - 2})

\frac{\partial}{\partial x} (x - \frac{2 y + 6 x}{13})

y = ln (\frac{x}{1 + x ^{2}})

\frac{4}{10} x^{2} d y - 13 d x = 0

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