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

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

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

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

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

Solution steps

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

Factor

3 x^{2} - 1 : - (- 3 x^{2} + 1)

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

Apply Integral Substitution

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

Substitute back

u = 3 x

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

Simplify

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

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

Add a constant to the solution

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

i = 2 \sum \infty \frac{5}{i ^{2} + i - 2}

x \to 5 lim (2 x + 3)

3 \cdot cos^{2} (t)

x^{^{'}} = 0.0333333 x (30 - x)

x \to - 1 lim (\frac{x ^{2} - 5}{x + 6})

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