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

The integral of (ln(1+x))/(1+x) is (ln^2(1+x))/2+C

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integral of (ln(1+x))/(1+x)

\int \frac{ln ( 1 + x )}{1 + x} d x

\frac{ln ^{2} ( 1 + x )}{2} + C

Solution steps

\int \frac{ln ( 1 + x )}{1 + x} d x

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

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

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{u ^{2}}{2}

Substitute back

u = ln (1 + x)

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

Add a constant to the solution

= \frac{ln ^{2} ( 1 + x )}{2} + C

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