What is the integral of (18)/(9x^2-x^4) ?

The integral of (18)/(9x^2-x^4) is 18(-1/(9x)+1/54 ln|x+3|-1/54 ln|x-3|)+C

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integral of (18)/(9x^2-x^4)

\int \frac{18}{9 x ^{2} - x ^{4}} d x

18 (- \frac{1}{9 x} + \frac{1}{54} ln ∣ x + 3 ∣ - \frac{1}{54} ln ∣ x - 3 ∣) + C

Solution steps

\int \frac{18}{9 x ^{2} - x ^{4}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 18 \cdot \int \frac{1}{9 x ^{2} - x ^{4}} d x

Take the partial fraction of

\frac{1}{9 x ^{2} - x ^{4}} : \frac{1}{9 x ^{2}} + \frac{1}{54 ( x + 3 )} - \frac{1}{54 ( x - 3 )}

= 18 \cdot \int \frac{1}{9 x ^{2}} + \frac{1}{54 ( x + 3 )} - \frac{1}{54 ( x - 3 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 18 (\int \frac{1}{9 x ^{2}} d x + \int \frac{1}{54 ( x + 3 )} d x - \int \frac{1}{54 ( x - 3 )} d x)

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

\int \frac{1}{54 ( x + 3 )} d x = \frac{1}{54} ln ∣ x + 3 ∣

\int \frac{1}{54 ( x - 3 )} d x = \frac{1}{54} ln ∣ x - 3 ∣

= 18 (- \frac{1}{9 x} + \frac{1}{54} ln ∣ x + 3 ∣ - \frac{1}{54} ln ∣ x - 3 ∣)

Add a constant to the solution

= 18 (- \frac{1}{9 x} + \frac{1}{54} ln ∣ x + 3 ∣ - \frac{1}{54} ln ∣ x - 3 ∣) + C

x \to 0 - lim (e^{\frac{x}{x + 1}})

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\int e^{2 x} 1 + e^{x} d x

x \to 0 lim (\frac{x - sin ( x )}{x})

What is the integral of (18)/(9x^2-x^4) ?
The integral of (18)/(9x^2-x^4) is 18(-1/(9x)+1/54 ln|x+3|-1/54 ln|x-3|)+C