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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Factor

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

= 2 \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

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

Apply Integral Substitution

= 2 (- \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

= 2 (- \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}

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

Substitute back

u = 2 x

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

Simplify

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

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

Add a constant to the solution

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

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