What is the integral of (6x-9)/(x^3+x) ?

The integral of (6x-9)/(x^3+x) is 6arctan(x)-9(ln|x|-1/2 ln|x^2+1|)+C

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integral of (6x-9)/(x^3+x)

\int \frac{6 x - 9}{x ^{3} + x} d x

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

Solution steps

\int \frac{6 x - 9}{x ^{3} + x} d x

Expand

\frac{6 x - 9}{x ^{3} + x} : \frac{6 x}{x ^{3} + x} - \frac{9}{x ^{3} + 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{6 x}{x ^{3} + x} d x - \int \frac{9}{x ^{3} + x} d x

\int \frac{6 x}{x ^{3} + x} d x = 6 arctan (x)

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

= 6 arctan (x) - 9 (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1)

Add a constant to the solution

= 6 arctan (x) - 9 (ln ∣ x ∣ - \frac{1}{2} ln x^{2} + 1) + C

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What is the integral of (6x-9)/(x^3+x) ?
The integral of (6x-9)/(x^3+x) is 6arctan(x)-9(ln|x|-1/2 ln|x^2+1|)+C