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

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

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

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

ln (2)

Decimal Notation

0.69314 \dots

Solution steps

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

y^{^{''}} + 4 y^{^{'}} + 5 y = 5 x - 1

\frac{\partial ^{2}}{\partial x \partial y} (y^{4} cos (2 x))

\frac{\partial}{\partial y} (16 - x^{2} - 4 y)

y^{^{''}} - y = sin (2 t)

f (x) = \frac{e ^{x}}{x + 3}

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