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

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

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

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

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

Solution steps

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

Long division

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

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

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

\int 1 d x = x

\int \frac{- 8 x + 8}{x ^{2} + 9} d x = - 4 ln \frac{x ^{2}}{9} + 1 + \frac{8}{3} arctan (\frac{x}{3})

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

Add a constant to the solution

= \frac{x ^{2}}{2} - x - 4 ln \frac{x ^{2}}{9} + 1 + \frac{8}{3} arctan (\frac{x}{3}) + C

\frac{x + 1}{x ^{2}}

x \to 2 lim (lo g_{4} (x - 2))

\frac{d y}{d x} + 3 x^{2} y = x^{2}

\frac{d y}{d x} = 10 x e^{5 y}

f (x) = \frac{2}{x}

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