What is the integral of 3/(4x^2-1) ?

The integral of 3/(4x^2-1) is -3/4 (ln|2x+1|-ln|2x-1|)+C

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integral of 3/(4x^2-1)

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

- \frac{3}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣) + C

Solution steps

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

Take the constant out:

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

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

Factor

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

= 3 \cdot \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

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

Apply Integral Substitution

= 3 (- \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

= 3 (- \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}

= 3 (- \frac{1}{2} (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = 2 x

= 3 (- \frac{1}{2} (\frac{ln ∣ 2 x + 1 ∣}{2} - \frac{ln ∣ 2 x - 1 ∣}{2}))

Simplify

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

= - \frac{3}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣)

Add a constant to the solution

= - \frac{3}{4} (ln ∣ 2 x + 1 ∣ - ln ∣ 2 x - 1 ∣) + C

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What is the integral of 3/(4x^2-1) ?
The integral of 3/(4x^2-1) is -3/4 (ln|2x+1|-ln|2x-1|)+C