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

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

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

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

2 ln (2) - 1

Solution steps

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

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

\int_{0}^{2} 2 x (x^{2} + 1)^{3} d x

\int_{- 2}^{- 1} x^{2} d x

\int_{7}^{49} \frac{1}{x ( ln ( x ) ) ^{9}} d x

\int_{1}^{2} 5 d x

\int_{2}^{3} 4^{4 x} d x

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