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

The integral from 0 to 2 of (ln(1+x))/(1+x) is (ln^2(3))/2

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

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

\frac{ln ^{2} ( 3 )}{2}

Decimal

0.60347 \dots

Solution steps

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

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

= \int_{0}^{2} ln (1 + x) \frac{1}{1 + x} d x

Apply u-substitution

= \int_{0}^{l n (3)} u d u

Apply the Power Rule

= [\frac{u ^{2}}{2}]_{0}^{l n (3)}

Compute the boundaries:

\frac{ln ^{2} ( 3 )}{2}

= \frac{ln ^{2} ( 3 )}{2}

What is the integral from 0 to 2 of (ln(1+x))/(1+x) ?
The integral from 0 to 2 of (ln(1+x))/(1+x) is (ln^2(3))/2