What is the integral of (2x-3)/(x(2x+1)(2x-1)) ?

The integral of (2x-3)/(x(2x+1)(2x-1)) is 3ln|x|-2ln|2x+1|-ln|2x-1|+C

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integral of (2x-3)/(x(2x+1)(2x-1))

\int \frac{2 x - 3}{x ( 2 x + 1 ) ( 2 x - 1 )} d x

3 ln ∣ x ∣ - 2 ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣ + C

Solution steps

\int \frac{2 x - 3}{x ( 2 x + 1 ) ( 2 x - 1 )} d x

Take the partial fraction of

\frac{2 x - 3}{x ( 2 x + 1 ) ( 2 x - 1 )} : \frac{3}{x} - \frac{4}{2 x + 1} - \frac{2}{2 x - 1}

= \int \frac{3}{x} - \frac{4}{2 x + 1} - \frac{2}{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{3}{x} d x - \int \frac{4}{2 x + 1} d x - \int \frac{2}{2 x - 1} d x

\int \frac{3}{x} d x = 3 ln ∣ x ∣

\int \frac{4}{2 x + 1} d x = 2 ln ∣ 2 x + 1 ∣

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

= 3 ln ∣ x ∣ - 2 ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣

Add a constant to the solution

= 3 ln ∣ x ∣ - 2 ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣ + C

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

\frac{d y}{d t} = y^{3} - y

\int \frac{1}{x 7 x - 1} d x

\frac{d}{d x} (e^{c o s (2 x)})

cos (π - 4 x)

What is the integral of (2x-3)/(x(2x+1)(2x-1)) ?
The integral of (2x-3)/(x(2x+1)(2x-1)) is 3ln|x|-2ln|2x+1|-ln|2x-1|+C