What is the integral of (x^2-6)/(x^2-9) ?

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

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integral of (x^2-6)/(x^2-9)

\int \frac{x ^{2} - 6}{x ^{2} - 9} d x

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

Solution steps

\int \frac{x ^{2} - 6}{x ^{2} - 9} d x

Apply Integral Substitution

= \int \frac{2 6 ( u ^{2} - 1 )}{2 u ^{2} - 3} d u

Take the constant out:

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

= 26 \cdot \int \frac{u ^{2} - 1}{2 u ^{2} - 3} d u

Long division

\frac{u ^{2} - 1}{2 u ^{2} - 3} : \frac{1}{2} + \frac{1}{2 ( 2 u ^{2} - 3 )}

= 26 \cdot \int \frac{1}{2} + \frac{1}{2 ( 2 u ^{2} - 3 )} d u

Apply the Sum Rule:

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

= 26 (\int \frac{1}{2} d u + \int \frac{1}{2 ( 2 u ^{2} - 3 )} d u)

\int \frac{1}{2} d u = \frac{1}{2} u

\int \frac{1}{2 ( 2 u ^{2} - 3 )} d u = - \frac{1}{4 6} (ln \frac{2}{3} u + 1 - ln \frac{2}{3} u - 1)

= 26 (\frac{1}{2} u - \frac{1}{4 6} (ln \frac{2}{3} u + 1 - ln \frac{2}{3} u - 1))

Substitute back

u = \frac{x}{6}

= 26 (\frac{1}{2} \cdot \frac{x}{6} - \frac{1}{4 6} (ln \frac{2}{3} \frac{x}{6} + 1 - ln \frac{2}{3} \frac{x}{6} - 1))

Simplify

26 (\frac{1}{2} \cdot \frac{x}{6} - \frac{1}{4 6} (ln \frac{2}{3} \frac{x}{6} + 1 - ln \frac{2}{3} \frac{x}{6} - 1)) : x - \frac{1}{2} ln \frac{x}{3} + 1 + \frac{1}{2} ln \frac{x}{3} - 1

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

Add a constant to the solution

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

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