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

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

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

\int ln (x + 1) - ln (x) d x

x (ln (x + 1) - ln (x)) + ln ∣ x + 1 ∣ + C

Solution steps

\int ln (x + 1) - ln (x) d x

Apply Integration By Parts

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

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

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

= x (ln (x + 1) - ln (x)) - (- ln ∣ x + 1 ∣)

Simplify

= x (ln (x + 1) - ln (x)) + ln ∣ x + 1 ∣

Add a constant to the solution

= x (ln (x + 1) - ln (x)) + ln ∣ x + 1 ∣ + C

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