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

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

integral of ln((x+2)/(x+1))

\int ln (\frac{x + 2}{x + 1}) d x

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

Solution steps

\int ln (\frac{x + 2}{x + 1}) d x

Apply u-substitution

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

Apply Integration By Parts

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

\int - \frac{1}{u - 1} d u = - ln ∣ u - 1 ∣

= u ln (\frac{u}{u - 1}) - (- ln ∣ u - 1 ∣)

Substitute back

u = x + 2

= (x + 2) ln (\frac{x + 2}{x + 2 - 1}) - (- ln ∣ x + 2 - 1 ∣)

Simplify

= ln (\frac{x + 2}{x + 1}) (x + 2) + ln ∣ x + 1 ∣

Add a constant to the solution

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

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d er i v a t i v e 2 1 + sin (2 t)

\int (x + 1) 8 x + 9 d x

t a y l or \frac{2 x}{( 1 + x ^{2} ) ^{2}}

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