What is the integral of (x^2-3x-1)/(x(x^2+4)) ?

The integral of (x^2-3x-1)/(x(x^2+4)) is 5/8 ln|x^2+4|-3/2 arctan(x/2)-1/4 ln|x|+C

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integral of (x^2-3x-1)/(x(x^2+4))

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

\frac{5}{8} ln x^{2} + 4 - \frac{3}{2} arctan (\frac{x}{2}) - \frac{1}{4} ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

= \frac{5}{8} ln x^{2} + 4 - \frac{3}{2} arctan (\frac{x}{2}) - \frac{1}{4} ln ∣ x ∣

Add a constant to the solution

= \frac{5}{8} ln x^{2} + 4 - \frac{3}{2} arctan (\frac{x}{2}) - \frac{1}{4} ln ∣ x ∣ + C

x^{- 2}

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What is the integral of (x^2-3x-1)/(x(x^2+4)) ?
The integral of (x^2-3x-1)/(x(x^2+4)) is 5/8 ln|x^2+4|-3/2 arctan(x/2)-1/4 ln|x|+C