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

= \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

\int x 1 + \frac{1}{4 x} d x

\int - 2 e^{t} d t

\int x^{4} (2 - x^{2}) d x

arccos (x)

\int \frac{3 x}{x ^{3}} d x

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