What is the integral of 1/(9x^2-4) ?

The integral of 1/(9x^2-4) is -1/12 (ln| 3/2 x+1|-ln| 3/2 x-1|)+C

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integral of 1/(9x^2-4)

\int \frac{1}{9 x ^{2} - 4} d x

- \frac{1}{12} (ln \frac{3}{2} x + 1 - ln \frac{3}{2} x - 1) + C

Solution steps

\int \frac{1}{9 x ^{2} - 4} d x

Apply Integral Substitution

= \int \frac{1}{6 ( 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}{6} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

u^{2} - 1 : - (- u^{2} + 1)

= \frac{1}{6} \cdot \int \frac{1}{- ( - 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}{6} (- \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}{6} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{3}{2} x

= \frac{1}{6} (- (\frac{ln \frac{3}{2} x + 1}{2} - \frac{ln \frac{3}{2} x - 1}{2}))

Simplify

\frac{1}{6} (- (\frac{ln \frac{3}{2} x + 1}{2} - \frac{ln \frac{3}{2} x - 1}{2})) : - \frac{1}{12} (ln \frac{3}{2} x + 1 - ln \frac{3}{2} x - 1)

= - \frac{1}{12} (ln \frac{3}{2} x + 1 - ln \frac{3}{2} x - 1)

Add a constant to the solution

= - \frac{1}{12} (ln \frac{3}{2} x + 1 - ln \frac{3}{2} x - 1) + C

\frac{\partial}{\partial x} (y^{3} + 3 x y - x^{3})

e^{- x^{3}}

\int \frac{x}{u}

\int y - x

y = x^{3}, 0 \leq x \leq 3

What is the integral of 1/(9x^2-4) ?
The integral of 1/(9x^2-4) is -1/12 (ln| 3/2 x+1|-ln| 3/2 x-1|)+C