What is the integral of (x^6-x^3+16)/(x^4+4x^2) ?

The integral of (x^6-x^3+16)/(x^4+4x^2) is (x^3)/3-4x-4/x-1/2 ln|(x^2)/4+1|+6arctan(x/2)+C

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integral of (x^6-x^3+16)/(x^4+4x^2)

\int \frac{x ^{6} - x ^{3} + 16}{x ^{4} + 4 x ^{2}} d x

\frac{x ^{3}}{3} - 4 x - \frac{4}{x} - \frac{1}{2} ln \frac{x ^{2}}{4} + 1 + 6 arctan (\frac{x}{2}) + C

Solution steps

\int \frac{x ^{6} - x ^{3} + 16}{x ^{4} + 4 x ^{2}} d x

Take the partial fraction of

\frac{x ^{6} - x ^{3} + 16}{x ^{4} + 4 x ^{2}} : x^{2} - 4 + \frac{4}{x ^{2}} + \frac{- x + 12}{x ^{2} + 4}

= \int x^{2} - 4 + \frac{4}{x ^{2}} + \frac{- x + 12}{x ^{2} + 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^{2} d x - \int 4 d x + \int \frac{4}{x ^{2}} d x + \int \frac{- x + 12}{x ^{2} + 4} d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 4 d x = 4 x

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

\int \frac{- x + 12}{x ^{2} + 4} d x = - \frac{1}{2} ln \frac{x ^{2}}{4} + 1 + 6 arctan (\frac{x}{2})

= \frac{x ^{3}}{3} - 4 x - \frac{4}{x} - \frac{1}{2} ln \frac{x ^{2}}{4} + 1 + 6 arctan (\frac{x}{2})

Add a constant to the solution

= \frac{x ^{3}}{3} - 4 x - \frac{4}{x} - \frac{1}{2} ln \frac{x ^{2}}{4} + 1 + 6 arctan (\frac{x}{2}) + C

t \frac{d y}{d t} = t^{3} + 15 t^{3} y

- 6 y^{^{'}} + 18 x^{2} = 0

(t^{3}) d x + (x t^{2} + x^{3}) d t = 0

\int cos (x) - cos^{3} (x) d x

What is the integral of (x^6-x^3+16)/(x^4+4x^2) ?
The integral of (x^6-x^3+16)/(x^4+4x^2) is (x^3)/3-4x-4/x-1/2 ln|(x^2)/4+1|+6arctan(x/2)+C