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

The integral of (x-1)/(x+2) is -3ln|x+2|+x+C

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

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

- 3 ln ∣ x + 2 ∣ + x + C

Solution steps

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

\frac{x - 1}{x + 2} = - \frac{3}{x + 2} + 1

= \int - \frac{3}{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{3}{x + 2} d x + \int 1 d x

\int \frac{3}{x + 2} d x = 3 ln ∣ x + 2 ∣

\int 1 d x = x

= - 3 ln ∣ x + 2 ∣ + x

Add a constant to the solution

= - 3 ln ∣ x + 2 ∣ + x + C

\int u \cdot sec^{2} (u) d u

f (x) = 8 x^{3}

\frac{d}{d x} ((sec^{2} (x)) d x)

f (t) = t^{6} sin (t)

cos (- x)

What is the integral of (x-1)/(x+2) ?
The integral of (x-1)/(x+2) is -3ln|x+2|+x+C