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

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

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

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

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

Solution steps

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

Expand

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

\int \frac{x ^{2}}{x ^{3} + 2 x} d x = \frac{1}{2} ln x^{2} + 2

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

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

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

Simplify

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

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

Add a constant to the solution

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

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

1 6^{4} 4 x^{4} + 4

6 y^{^{'}} = x^{2}

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

y^{^{'}} = \frac{2 x ^{4} + y ^{4}}{x y ^{3}}

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