What is the integral of 1/(9x^2-144) ?

The integral of 1/(9x^2-144) is -1/72 (ln| x/4+1|-ln| x/4-1|)+C

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integral of 1/(9x^2-144)

\int \frac{1}{9 x ^{2} - 144} d x

- \frac{1}{72} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1) + C

Solution steps

\int \frac{1}{9 x ^{2} - 144} d x

Apply Integral Substitution

= \int \frac{1}{36 ( 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}{36} \cdot \int \frac{1}{u ^{2} - 1} d u

Factor

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

= \frac{1}{36} \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}{36} (- \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}{36} (- (\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}))

Substitute back

u = \frac{3}{12} x

= \frac{1}{36} (- (\frac{ln \frac{3}{12} x + 1}{2} - \frac{ln \frac{3}{12} x - 1}{2}))

Simplify

\frac{1}{36} (- (\frac{ln \frac{3}{12} x + 1}{2} - \frac{ln \frac{3}{12} x - 1}{2})) : - \frac{1}{72} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1)

= - \frac{1}{72} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1)

Add a constant to the solution

= - \frac{1}{72} (ln \frac{x}{4} + 1 - ln \frac{x}{4} - 1) + C

y = x + 2, y = \frac{1}{x + 1}, x = 2

\frac{d}{d x} (2 cot (x))

\frac{\partial}{\partial x} (2 x^{2} - 2 x y^{2})

\int \frac{x + 3}{x - 1} d x

\frac{d}{d x} (arcsin (\frac{x}{π}))

What is the integral of 1/(9x^2-144) ?
The integral of 1/(9x^2-144) is -1/72 (ln| x/4+1|-ln| x/4-1|)+C