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

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

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

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

2 ln (2) - 1

Solution steps

\int_{0}^{1} ln (y + 1) 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_{- 3}^{2} 4 sin (x^{2}) - 1 d x

\int_{- 2}^{0} 6 x^{2} (2 x^{3} + 5)^{2} d x

\int_{- 1}^{1} 4 1 - x^{2} d x

\int_{0}^{1} 2 x - x^{2} - x^{2} d x

\int_{a}^{b} x^{2} \frac{1}{b - a} d x

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