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

y^{^{'}} (t) = (y + 1) \cdot (y - 3), y (0) = 1

\frac{d}{d x} (lo g_{10} (sin (x)))

x \to - 4 lim (5 x + 20 - 8)

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

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

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