What is the integral of (9x^2-8x+20)/(x^3+4x) ?

The integral of (9x^2-8x+20)/(x^3+4x) is 2ln|x^2+4|-4arctan(x/2)+5ln|x|+C

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integral of (9x^2-8x+20)/(x^3+4x)

\int \frac{9 x ^{2} - 8 x + 20}{x ^{3} + 4 x} d x

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

Solution steps

\int \frac{9 x ^{2} - 8 x + 20}{x ^{3} + 4 x} d x

Expand

\frac{9 x ^{2} - 8 x + 20}{x ^{3} + 4 x} : \frac{9 x ^{2}}{x ^{3} + 4 x} - \frac{8 x}{x ^{3} + 4 x} + \frac{20}{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{9 x ^{2}}{x ^{3} + 4 x} d x - \int \frac{8 x}{x ^{3} + 4 x} d x + \int \frac{20}{x ^{3} + 4 x} d x

\int \frac{9 x ^{2}}{x ^{3} + 4 x} d x = \frac{9}{2} ln x^{2} + 4

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

\int \frac{20}{x ^{3} + 4 x} d x = 20 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

= \frac{9}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 20 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4)

Simplify

\frac{9}{2} ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 20 (\frac{1}{4} ln ∣ x ∣ - \frac{1}{8} ln x^{2} + 4) : 2 ln x^{2} + 4 - 4 arctan (\frac{x}{2}) + 5 ln ∣ x ∣

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

Add a constant to the solution

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

\frac{75 x}{x ^{2} + 100}

1 - 2 x

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

y^{^{''}} - 9 y = 0, y (0) = 6

x \to 54 lim (\frac{x ^{2}}{6} - \frac{9}{x})

What is the integral of (9x^2-8x+20)/(x^3+4x) ?
The integral of (9x^2-8x+20)/(x^3+4x) is 2ln|x^2+4|-4arctan(x/2)+5ln|x|+C