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

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

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

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

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

Solution steps

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

Expand

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

\int \frac{x ^{2}}{x ^{3} + 4 x} d x = \frac{1}{2} ln x^{2} + 4

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

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

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

Add a constant to the solution

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

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