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

The integral of xln((x+2)/(x+1)) is 1/2 x^2ln((x+2)/(x+1))+1/2 (x-4ln|x+2|+ln|x+1|)+C

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

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

\frac{1}{2} x^{2} ln (\frac{x + 2}{x + 1}) + \frac{1}{2} (x - 4 ln ∣ x + 2 ∣ + ln ∣ x + 1 ∣) + C

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{x ^{2}}{2 ( x + 2 ) ( x + 1 )} d x = - \frac{1}{2} (x - 4 ln ∣ x + 2 ∣ + ln ∣ x + 1 ∣)

= \frac{1}{2} x^{2} ln (\frac{x + 2}{x + 1}) - (- \frac{1}{2} (x - 4 ln ∣ x + 2 ∣ + ln ∣ x + 1 ∣))

Simplify

= \frac{1}{2} x^{2} ln (\frac{x + 2}{x + 1}) + \frac{1}{2} (x - 4 ln ∣ x + 2 ∣ + ln ∣ x + 1 ∣)

Add a constant to the solution

= \frac{1}{2} x^{2} ln (\frac{x + 2}{x + 1}) + \frac{1}{2} (x - 4 ln ∣ x + 2 ∣ + ln ∣ x + 1 ∣) + C

\int \frac{4 x}{y} d x

\frac{d}{d x} (x \cdot e^{- x})

\frac{d y}{d x} = 9 x e^{y}, y (0) = 0

f (x) = 2 x^{2} - 1, at x = - 1

\frac{d y}{d x} = x^{2} y - y + x^{2} - 1

What is the integral of xln((x+2)/(x+1)) ?
The integral of xln((x+2)/(x+1)) is 1/2 x^2ln((x+2)/(x+1))+1/2 (x-4ln|x+2|+ln|x+1|)+C