What is the integral of (x-1)/(x+4) ?

The integral of (x-1)/(x+4) is -5ln|x+4|+x+C

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integral of (x-1)/(x+4)

\int \frac{x - 1}{x + 4} d x

- 5 ln ∣ x + 4 ∣ + x + C

Solution steps

\int \frac{x - 1}{x + 4} d x

\frac{x - 1}{x + 4} = - \frac{5}{x + 4} + 1

= \int - \frac{5}{x + 4} + 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{5}{x + 4} d x + \int 1 d x

\int \frac{5}{x + 4} d x = 5 ln ∣ x + 4 ∣

\int 1 d x = x

= - 5 ln ∣ x + 4 ∣ + x

Add a constant to the solution

= - 5 ln ∣ x + 4 ∣ + x + C

\int \frac{2 x - 4}{x ^{3} + 4 x ^{2}} d x

\frac{d}{d x} (3 x^{- \frac{5}{3}})

\frac{d}{d x} (cos (sin (2 x)))

\frac{d}{d x} (a cos (7 x) + b sin (7 x))

n = 1 \sum \infty \frac{2 ^{- n}}{5 ^{- n}}

What is the integral of (x-1)/(x+4) ?
The integral of (x-1)/(x+4) is -5ln|x+4|+x+C