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

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

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

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

- ln ∣ x + 2 ∣ + x + C

Solution steps

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

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

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

\int \frac{1}{x + 2} d x = ln ∣ x + 2 ∣

\int 1 d x = x

= - ln ∣ x + 2 ∣ + x

Add a constant to the solution

= - ln ∣ x + 2 ∣ + x + C

\int \frac{x}{x + 20} d x

(x + x) (\frac{d y}{d x}) = y + y

f (x) = cos (x), at (0)

y = \frac{( 2 x + 1 ) ( x ^{3} + 3 )}{x}

4 x^{2} - 2 x + 3

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