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

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

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

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

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

Decimal

0.24022 \dots

Solution steps

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

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

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

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