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

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

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

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

\frac{ln ^{2} ( 2 )}{2}

Decimal

0.24022 \dots

Solution steps

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

\frac{ln ( x + 1 )}{x + 1} = ln (x + 1) \frac{1}{x + 1}

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

Apply u-substitution

: \int_{0}^{l n (2)} u d u

= \int_{0}^{l n (2)} u d u

Apply the Power Rule

: [\frac{u ^{2}}{2}]_{0}^{l n (2)}

= [\frac{u ^{2}}{2}]_{0}^{l n (2)}

Compute the boundaries:

\frac{ln ^{2} ( 2 )}{2}

= \frac{ln ^{2} ( 2 )}{2}

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