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

The integral of (4-x)/(x^3+9x) is 4(1/9 ln|x|-1/18 ln|x^2+9|)-1/3 arctan(x/3)+C

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

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

4 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9) - \frac{1}{3} arctan (\frac{x}{3}) + C

Solution steps

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

Expand

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

\int \frac{4}{x ^{3} + 9 x} d x = 4 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9)

\int \frac{x}{x ^{3} + 9 x} d x = \frac{1}{3} arctan (\frac{x}{3})

= 4 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9) - \frac{1}{3} arctan (\frac{x}{3})

Add a constant to the solution

= 4 (\frac{1}{9} ln ∣ x ∣ - \frac{1}{18} ln x^{2} + 9) - \frac{1}{3} arctan (\frac{x}{3}) + C

What is the integral of (4-x)/(x^3+9x) ?
The integral of (4-x)/(x^3+9x) is 4(1/9 ln|x|-1/18 ln|x^2+9|)-1/3 arctan(x/3)+C