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

The integral of ln(x/(x+1)) is xln(x/(x+1))-ln|x+1|+C

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

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

x ln (\frac{x}{x + 1}) - ln ∣ x + 1 ∣ + C

Solution steps

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

Apply Integration By Parts

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

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

\int \frac{1}{x + 1} d x = ln ∣ x + 1 ∣

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

Add a constant to the solution

= x ln (\frac{x}{x + 1}) - ln ∣ x + 1 ∣ + C

\int_{0}^{4} x^{2} - 4 x + 3 d x

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9 x

f (x) = 4 sin (x) + 7 cos (x)

\frac{\partial}{\partial x} (x^{4} sin (x y^{3}))

What is the integral of ln(x/(x+1)) ?
The integral of ln(x/(x+1)) is xln(x/(x+1))-ln|x+1|+C