What is the integral of (2x-1)/(4x^2-1) ?

The integral of (2x-1)/(4x^2-1) is 1/4 ln|4x^2-1|+1/4 (ln|2x+1|-ln|2x-1|)+C

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integral of (2x-1)/(4x^2-1)

\int \frac{2 x - 1}{4 x ^{2} - 1} d x

\frac{1}{4} ln 4 x^{2} - 1 + \frac{1}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣) + C

Solution steps

\int \frac{2 x - 1}{4 x ^{2} - 1} d x

Expand

\frac{2 x - 1}{4 x ^{2} - 1} : \frac{2 x}{4 x ^{2} - 1} - \frac{1}{4 x ^{2} - 1}

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{2 x}{4 x ^{2} - 1} d x - \int \frac{1}{4 x ^{2} - 1} d x

\int \frac{2 x}{4 x ^{2} - 1} d x = \frac{1}{4} ln 4 x^{2} - 1

\int \frac{1}{4 x ^{2} - 1} d x = - \frac{1}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣)

= \frac{1}{4} ln 4 x^{2} - 1 - (- \frac{1}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣))

Simplify

= \frac{1}{4} ln 4 x^{2} - 1 + \frac{1}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣)

Add a constant to the solution

= \frac{1}{4} ln 4 x^{2} - 1 + \frac{1}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣) + C

\int_{0}^{4} π (4 y^{2} - \frac{y ^{4}}{4}) d y

- 4 t (6 t^{4} - 1)^{7}

\int_{0}^{1} x^{2} (1 + 2 x^{3})^{5} d x

y = \frac{4 x ^{2} + 6 x + 4}{x}

\frac{\partial}{\partial y} (5 x y^{2})

What is the integral of (2x-1)/(4x^2-1) ?
The integral of (2x-1)/(4x^2-1) is 1/4 ln|4x^2-1|+1/4 (ln|2x+1|-ln|2x-1|)+C