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

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

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

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

x - 1 + 5 ln ∣ x - 1 ∣ + C

Solution steps

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

Expand

\frac{x + 4}{x - 1} : \frac{x}{x - 1} + \frac{4}{x - 1}

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

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

\int \frac{4}{x - 1} d x = 4 ln ∣ x - 1 ∣

= x - 1 + ln ∣ x - 1 ∣ + 4 ln ∣ x - 1 ∣

Simplify

= x - 1 + 5 ln ∣ x - 1 ∣

Add a constant to the solution

= x - 1 + 5 ln ∣ x - 1 ∣ + C

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