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

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

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

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

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

Solution steps

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

Factor

4 x^{2} - 1 : - (- 4 x^{2} + 1)

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply Integral Substitution

= - \int \frac{1}{2 ( - u ^{2} + 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{2} \cdot \int \frac{1}{- u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= - \frac{1}{2} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = 2 x

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (sin (6 x y + 7 yz + 9 x z))

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

\int (4 x - \frac{2}{x} + 5 sin (x)) d x

x \to \infty lim ((4)^{x})

f^{^{'}} (x) = sec (2 x) tan (2 x) - \frac{4}{x ^{2}}

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