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

\frac{\partial}{\partial x} (x ln (x + 2 y))

\frac{1}{s ^{2} + 6 s + 34}

1 - e^{- t}

cos (x), sin (x), [0, \frac{π}{4}]

\int \frac{7}{( x ^{2} + 1 ) ^{\frac{3}{2}}} 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