What is the integral of 1/(9x^2-1) ?

The integral of 1/(9x^2-1) is -1/6 (ln|3x+1|-ln|3x-1|)+C

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integral of 1/(9x^2-1)

\int \frac{1}{9 x ^{2} - 1} d x

- \frac{1}{6} (ln ∣ 3 x + 1 ∣ - ln ∣ 3 x - 1 ∣) + C

Solution steps

\int \frac{1}{9 x ^{2} - 1} d x

Factor

9 x^{2} - 1 : - (- 9 x^{2} + 1)

= \int \frac{1}{- ( - 9 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}{- 9 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}{6} (ln ∣ 3 x + 1 ∣ - ln ∣ 3 x - 1 ∣)

= - \frac{1}{6} (ln ∣ 3 x + 1 ∣ - ln ∣ 3 x - 1 ∣)

Add a constant to the solution

= - \frac{1}{6} (ln ∣ 3 x + 1 ∣ - ln ∣ 3 x - 1 ∣) + C

What is the integral of 1/(9x^2-1) ?
The integral of 1/(9x^2-1) is -1/6 (ln|3x+1|-ln|3x-1|)+C