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

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

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

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

- ln ∣ x ∣ - \frac{1}{x} + \frac{2}{3} ln ∣ x + 3 ∣ + \frac{1}{3} ln ∣ x - 3 ∣ + C

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

\frac{x - 1}{x ^{2} ( x ^{2} - 9 )} : - \frac{1}{9 x} + \frac{1}{9 x ^{2}} + \frac{2}{27 ( x + 3 )} + \frac{1}{27 ( x - 3 )}

= 9 \cdot \int - \frac{1}{9 x} + \frac{1}{9 x ^{2}} + \frac{2}{27 ( x + 3 )} + \frac{1}{27 ( 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

= 9 (- \int \frac{1}{9 x} d x + \int \frac{1}{9 x ^{2}} d x + \int \frac{2}{27 ( x + 3 )} d x + \int \frac{1}{27 ( x - 3 )} d x)

\int \frac{1}{9 x} d x = \frac{1}{9} ln ∣ x ∣

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

\int \frac{2}{27 ( x + 3 )} d x = \frac{2}{27} ln ∣ x + 3 ∣

\int \frac{1}{27 ( x - 3 )} d x = \frac{1}{27} ln ∣ x - 3 ∣

= 9 (- \frac{1}{9} ln ∣ x ∣ - \frac{1}{9 x} + \frac{2}{27} ln ∣ x + 3 ∣ + \frac{1}{27} ln ∣ x - 3 ∣)

Simplify

9 (- \frac{1}{9} ln ∣ x ∣ - \frac{1}{9 x} + \frac{2}{27} ln ∣ x + 3 ∣ + \frac{1}{27} ln ∣ x - 3 ∣) : - ln ∣ x ∣ - \frac{1}{x} + \frac{2}{3} ln ∣ x + 3 ∣ + \frac{1}{3} ln ∣ x - 3 ∣

= - ln ∣ x ∣ - \frac{1}{x} + \frac{2}{3} ln ∣ x + 3 ∣ + \frac{1}{3} ln ∣ x - 3 ∣

Add a constant to the solution

= - ln ∣ x ∣ - \frac{1}{x} + \frac{2}{3} ln ∣ x + 3 ∣ + \frac{1}{3} ln ∣ x - 3 ∣ + C

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