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

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

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

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

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

Solution steps

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

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

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

Apply u-substitution

= \int u d u

Apply the Power Rule

= \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

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