What is the integral of (x^4-1)/(x^3-x^2+9x-9) ?

The integral of (x^4-1)/(x^3-x^2+9x-9) is (x^2)/2+x-8(1/2 ln|x^2+9|+1/3 arctan(x/3))+C

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integral of (x^4-1)/(x^3-x^2+9x-9)

\int \frac{x ^{4} - 1}{x ^{3} - x ^{2} + 9 x - 9} d x

\frac{x ^{2}}{2} + x - 8 (\frac{1}{2} ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3})) + C

Solution steps

\int \frac{x ^{4} - 1}{x ^{3} - x ^{2} + 9 x - 9} d x

Long division

\frac{x ^{4} - 1}{x ^{3} - x ^{2} + 9 x - 9} : x + 1 - \frac{8 ( x + 1 )}{x ^{2} + 9}

= \int x + 1 - \frac{8 ( x + 1 )}{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

= \int x d x + \int 1 d x - \int \frac{8 ( x + 1 )}{x ^{2} + 9} d x

\int x d x = \frac{x ^{2}}{2}

\int 1 d x = x

\int \frac{8 ( x + 1 )}{x ^{2} + 9} d x = 8 (\frac{1}{2} ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3}))

= \frac{x ^{2}}{2} + x - 8 (\frac{1}{2} ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3}))

Add a constant to the solution

= \frac{x ^{2}}{2} + x - 8 (\frac{1}{2} ln x^{2} + 9 + \frac{1}{3} arctan (\frac{x}{3})) + C

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What is the integral of (x^4-1)/(x^3-x^2+9x-9) ?
The integral of (x^4-1)/(x^3-x^2+9x-9) is (x^2)/2+x-8(1/2 ln|x^2+9|+1/3 arctan(x/3))+C