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

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

\int 6 x + 3 d x

n = 1 \sum \infty \frac{1}{e ^{n^{3}}}

\int \frac{1}{( 2 x - 1 ) ^{3}} d x

x \to - 6 lim (\frac{x ^{2} + 6}{x - 6})

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