What is the integral of (x+1)/(x-5) ?

The integral of (x+1)/(x-5) is 6ln|x-5|+x+C

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integral of (x+1)/(x-5)

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

6 ln ∣ x - 5 ∣ + x + C

Solution steps

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

\frac{x + 1}{x - 5} = \frac{6}{x - 5} + 1

= \int \frac{6}{x - 5} + 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{6}{x - 5} d x + \int 1 d x

\int \frac{6}{x - 5} d x = 6 ln ∣ x - 5 ∣

\int 1 d x = x

= 6 ln ∣ x - 5 ∣ + x

Add a constant to the solution

= 6 ln ∣ x - 5 ∣ + x + C

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What is the integral of (x+1)/(x-5) ?
The integral of (x+1)/(x-5) is 6ln|x-5|+x+C