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

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

\int \frac{35}{1 - cos ( 5 x )} d x

\frac{d}{d x} (x^{5} + x + 1)

\frac{d}{d x} (\frac{( ( 0.06 x ^{2} ) + 17 x + 340 )}{x})

l a pl a ce t r an s f or m 2 e^{\frac{t}{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