What is the integral of [ 1/(2x-1)-1/(2x+1)] ?

The integral of [ 1/(2x-1)-1/(2x+1)] is 1/2 ln|2x-1|-1/2 ln|2x+1|+C

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integral of [ 1/(2x-1)-1/(2x+1)]

\int [\frac{1}{2 x - 1} - \frac{1}{2 x + 1}] d x

\frac{1}{2} ln ∣ 2 x - 1 ∣ - \frac{1}{2} ln ∣ 2 x + 1 ∣ + C

Solution steps

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

\int \frac{1}{2 x - 1} d x = \frac{1}{2} ln ∣ 2 x - 1 ∣

\int \frac{1}{2 x + 1} d x = \frac{1}{2} ln ∣ 2 x + 1 ∣

= \frac{1}{2} ln ∣ 2 x - 1 ∣ - \frac{1}{2} ln ∣ 2 x + 1 ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 2 x - 1 ∣ - \frac{1}{2} ln ∣ 2 x + 1 ∣ + C

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What is the integral of [ 1/(2x-1)-1/(2x+1)] ?
The integral of [ 1/(2x-1)-1/(2x+1)] is 1/2 ln|2x-1|-1/2 ln|2x+1|+C