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

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

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

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{1}{3 ( 1 - u ^{2} )} 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}{1 - u ^{2}} d u

Use the common integral:

\int \frac{1}{1 - u ^{2}} 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

\int 2 cos (x^{2}) d x

\frac{d}{d x} (\frac{2 x ^{2} + 8 x + 8}{x})

f (x) = (2 e^{x} - 3 x^{5})

\int \frac{1}{4 t ^{2} - 4 t + 5} d t

\frac{\partial}{\partial x} (- 4 y^{3} sin (4 x))

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