What is the integral of (x^3-8)/(x^2+8) ?

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

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integral of (x^3-8)/(x^2+8)

\int \frac{x ^{3} - 8}{x ^{2} + 8} d x

\frac{1}{2} (x^{2} + 8 - 8 ln x^{2} + 8) - 22 arctan (\frac{x}{2 2}) + C

Solution steps

\int \frac{x ^{3} - 8}{x ^{2} + 8} d x

Expand

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x ^{3}}{x ^{2} + 8} d x - \int \frac{8}{x ^{2} + 8} d x

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

\int \frac{8}{x ^{2} + 8} d x = 22 arctan (\frac{x}{2 2})

= \frac{1}{2} (x^{2} + 8 - 8 ln x^{2} + 8) - 22 arctan (\frac{x}{2 2})

Add a constant to the solution

= \frac{1}{2} (x^{2} + 8 - 8 ln x^{2} + 8) - 22 arctan (\frac{x}{2 2}) + C

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