What is the integral of 2/(x^3-2x^2+9x-18) ?

The integral of 2/(x^3-2x^2+9x-18) is 2(1/13 ln|x-2|+1/39 (-3/2 ln|(x^2)/9+1|-2arctan(x/3)))+C

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integral of 2/(x^3-2x^2+9x-18)

\int \frac{2}{x ^{3} - 2 x ^{2} + 9 x - 18} d x

2 (\frac{1}{13} ln ∣ x - 2 ∣ + \frac{1}{39} (- \frac{3}{2} ln \frac{x ^{2}}{9} + 1 - 2 arctan (\frac{x}{3}))) + C

Solution steps

\int \frac{2}{x ^{3} - 2 x ^{2} + 9 x - 18} d x

Take the constant out:

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

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

Take the partial fraction of

\frac{1}{x ^{3} - 2 x ^{2} + 9 x - 18} : \frac{1}{13 ( x - 2 )} + \frac{- x - 2}{13 ( x ^{2} + 9 )}

= 2 \cdot \int \frac{1}{13 ( x - 2 )} + \frac{- x - 2}{13 ( 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

= 2 (\int \frac{1}{13 ( x - 2 )} d x + \int \frac{- x - 2}{13 ( x ^{2} + 9 )} d x)

\int \frac{1}{13 ( x - 2 )} d x = \frac{1}{13} ln ∣ x - 2 ∣

\int \frac{- x - 2}{13 ( x ^{2} + 9 )} d x = \frac{1}{39} (- \frac{3}{2} ln \frac{x ^{2}}{9} + 1 - 2 arctan (\frac{x}{3}))

= 2 (\frac{1}{13} ln ∣ x - 2 ∣ + \frac{1}{39} (- \frac{3}{2} ln \frac{x ^{2}}{9} + 1 - 2 arctan (\frac{x}{3})))

Add a constant to the solution

= 2 (\frac{1}{13} ln ∣ x - 2 ∣ + \frac{1}{39} (- \frac{3}{2} ln \frac{x ^{2}}{9} + 1 - 2 arctan (\frac{x}{3}))) + C

3 e^{- x} + e^{4 x}

\frac{d}{d x} (\frac{4 x + 5}{4 x - 5})

\int (\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}) d x

\int \frac{6 x}{( 4 - 3 x ^{2} ) ^{4}} d x

\frac{d}{d x} (ln (\frac{x ^{2} + 4}{x ^{2} + 1}))

What is the integral of 2/(x^3-2x^2+9x-18) ?
The integral of 2/(x^3-2x^2+9x-18) is 2(1/13 ln|x-2|+1/39 (-3/2 ln|(x^2)/9+1|-2arctan(x/3)))+C