What is the integral of 1/x-1/(x+6) ?

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

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integral of 1/x-1/(x+6)

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

ln ∣ x ∣ - ln ∣ x + 6 ∣ + C

Solution steps

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

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{1}{x + 6} d x = ln ∣ x + 6 ∣

= ln ∣ x ∣ - ln ∣ x + 6 ∣

Add a constant to the solution

= ln ∣ x ∣ - ln ∣ x + 6 ∣ + C

\int y 2 + 10 y - 25 y^{2} d y

\frac{( x + 1 ) ^{2}}{( x - 1 ) ( x + 2 )}

x \to \infty lim (x (x - x^{2} - 4))

\int 10 x^{- \frac{1}{2}} + 1 d x

2 \frac{1}{x}, at x = 3

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