What is the integral of (x^2-5)/(x^2(x^2-9)) ?

The integral of (x^2-5)/(x^2(x^2-9)) is

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integral of (x^2-5)/(x^2(x^2-9))

\int \frac{x ^{2} - 5}{x ^{2} ( x ^{2} - 9 )} d x

- \frac{1}{6} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1) - 5 (\frac{1}{9 x} - \frac{1}{54} ln ∣ x + 3 ∣ + \frac{1}{54} ln ∣ x - 3 ∣) + C

Solution steps

\int \frac{x ^{2} - 5}{x ^{2} ( x ^{2} - 9 )} d x

Expand

\frac{x ^{2} - 5}{x ^{2} ( x ^{2} - 9 )} : \frac{x ^{2}}{x ^{2} ( x ^{2} - 9 )} - \frac{5}{x ^{2} ( x ^{2} - 9 )}

Apply the Sum Rule:

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

= \int \frac{x ^{2}}{x ^{2} ( x ^{2} - 9 )} d x - \int \frac{5}{x ^{2} ( x ^{2} - 9 )} d x

\int \frac{x ^{2}}{x ^{2} ( x ^{2} - 9 )} d x = - \frac{1}{6} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1)

\int \frac{5}{x ^{2} ( x ^{2} - 9 )} d x = 5 (\frac{1}{9 x} - \frac{1}{54} ln ∣ x + 3 ∣ + \frac{1}{54} ln ∣ x - 3 ∣)

= - \frac{1}{6} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1) - 5 (\frac{1}{9 x} - \frac{1}{54} ln ∣ x + 3 ∣ + \frac{1}{54} ln ∣ x - 3 ∣)

Add a constant to the solution

= - \frac{1}{6} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1) - 5 (\frac{1}{9 x} - \frac{1}{54} ln ∣ x + 3 ∣ + \frac{1}{54} ln ∣ x - 3 ∣) + C

\int \frac{1}{2 x ^{2} + 3 x - 2} d x

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

\int \frac{3 - x ^{2}}{x ^{3} - x} d x

\int \frac{1 + tan ( x )}{cos ^{2} ( x )} d x

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

What is the integral of (x^2-5)/(x^2(x^2-9)) ?
The integral of (x^2-5)/(x^2(x^2-9)) is