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

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

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

\int \frac{x + 2}{x + 1} d x

ln ∣ x + 1 ∣ + x + C

Solution steps

\int \frac{x + 2}{x + 1} d x

\frac{x + 2}{x + 1} = \frac{1}{x + 1} + 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d x = x

= ln ∣ x + 1 ∣ + x

Add a constant to the solution

= ln ∣ x + 1 ∣ + x + C

\frac{d}{d x} (2 (3 x^{2} + x - 3)^{2})

\int \frac{x ^{2} - 4}{x ^{2}} d x

x \to 1 lim (\frac{x - 2}{x ^{2} - 4})

y = \frac{4 x - 9}{2 x + 3}, x = 2

(2 x - 2)^{4} (x^{2} + x + 1)^{5}

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