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

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

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

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

2 ln (2) - 1

Decimal

0.38629 \dots

Solution steps

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

Apply u-substitution

: \int_{1}^{2} ln (u) d u

= \int_{1}^{2} ln (u) d u

Apply Integration By Parts

: [u ln (u) - \int 1 d u]_{1}^{2}

= [u ln (u) - \int 1 d u]_{1}^{2}

\int 1 d u = u

= [u ln (u) - u]_{1}^{2}

Compute the boundaries:

2 ln (2) - 1

= 2 ln (2) - 1

\frac{d}{d x} (\frac{10 x + 4 x ^{2}}{( 5 + 4 x ) ^{2}})

\frac{d}{d x} ((2 x + 5)^{3} \cdot (3 x - 1)^{4})

\int (- 2^{3 - 2 x}) d x

\int \frac{y}{x} d x

\frac{\partial}{\partial y} (cos (x) + sin (y))

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