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

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

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

\int (\frac{1}{x} - \frac{1}{x + 1}) d x

ln ∣ x ∣ - ln ∣ x + 1 ∣ + C

Solution steps

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ - ln ∣ x + 1 ∣

Add a constant to the solution

= ln ∣ x ∣ - ln ∣ x + 1 ∣ + C

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

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\int x^{5} 14 + x^{6} d x

\int \frac{2}{x ^{2} ( x + 1 )} d x

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