What is the integral of (10)/(x^3-x^2+9x-9) ?

The integral of (10)/(x^3-x^2+9x-9) is ln|x-1|-1/2 ln|x^2+9|-1/3 arctan(x/3)+C

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integral of (10)/(x^3-x^2+9x-9)

\int \frac{10}{x ^{3} - x ^{2} + 9 x - 9} d x

ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}) + C

Solution steps

\int \frac{10}{x ^{3} - x ^{2} + 9 x - 9} d x

Take the constant out:

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

= 10 \cdot \int \frac{1}{x ^{3} - x ^{2} + 9 x - 9} d x

Take the partial fraction of

\frac{1}{x ^{3} - x ^{2} + 9 x - 9} : \frac{1}{10 ( x - 1 )} + \frac{- x - 1}{10 ( x ^{2} + 9 )}

= 10 \cdot \int \frac{1}{10 ( x - 1 )} + \frac{- x - 1}{10 ( x ^{2} + 9 )} 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

= 10 (\int \frac{1}{10 ( x - 1 )} d x + \int \frac{- x - 1}{10 ( x ^{2} + 9 )} d x)

\int \frac{1}{10 ( x - 1 )} d x = \frac{1}{10} ln ∣ x - 1 ∣

\int \frac{- x - 1}{10 ( x ^{2} + 9 )} d x = \frac{1}{10} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}))

= 10 (\frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{10} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3})))

Simplify

10 (\frac{1}{10} ln ∣ x - 1 ∣ + \frac{1}{10} (- \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3}))) : ln ∣ x - 1 ∣ - \frac{1}{2} ln x^{2} + 9 - \frac{1}{3} arctan (\frac{x}{3})

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

Add a constant to the solution

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

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What is the integral of (10)/(x^3-x^2+9x-9) ?
The integral of (10)/(x^3-x^2+9x-9) is ln|x-1|-1/2 ln|x^2+9|-1/3 arctan(x/3)+C