What is the integral of (x^2+3x+5)/(x^3+8) ?

The integral of (x^2+3x+5)/(x^3+8) is

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integral of (x^2+3x+5)/(x^3+8)

\int \frac{x ^{2} + 3 x + 5}{x ^{3} + 8} d x

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

Solution steps

\int \frac{x ^{2} + 3 x + 5}{x ^{3} + 8} d x

Take the partial fraction of

\frac{x ^{2} + 3 x + 5}{x ^{3} + 8} : \frac{1}{4 ( x + 2 )} + \frac{3 x + 8}{4 ( x ^{2} - 2 x + 4 )}

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

\int \frac{1}{4 ( x + 2 )} d x = \frac{1}{4} ln ∣ x + 2 ∣

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

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

Add a constant to the solution

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

\frac{d ^{4}}{d x ^{4}} (e^{2 x})

2 x + 8, x, [0, 9]

\int 5 ln (x^{2} - 1) d x

\int (l - x)^{2} d x

\frac{\partial}{\partial x} (e^{x y^{2}} y^{2})

What is the integral of (x^2+3x+5)/(x^3+8) ?
The integral of (x^2+3x+5)/(x^3+8) is