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

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

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

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

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

Solution steps

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

Long division

\frac{x ^{3}}{x ^{2} - 2 x + 4} : x + 2 - \frac{8}{x ^{2} - 2 x + 4}

= \int x + 2 - \frac{8}{x ^{2} - 2 x + 4} 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 2 d x - \int \frac{8}{x ^{2} - 2 x + 4} d x

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

\int 2 d x = 2 x

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

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

Add a constant to the solution

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

\int \frac{sec ( x )}{x} d x

y^{^{''}} - 2 y^{^{'}} - 3 y = - 3 t e^{- t}

x \to 1 lim (x - 2)

x \to \infty lim (\frac{2 ^{x + 1}}{3 ^{x - 1}})

\int x^{2} sin (4 x) d x

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