What is the integral of (3x+1)/(x-x^3) ?

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

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integral of (3x+1)/(x-x^3)

\int \frac{3 x + 1}{x - x ^{3}} d x

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

Solution steps

\int \frac{3 x + 1}{x - x ^{3}} d x

Expand

\frac{3 x + 1}{x - x ^{3}} : \frac{3 x}{x - x ^{3}} + \frac{1}{x - x ^{3}}

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{3 x}{x - x ^{3}} d x + \int \frac{1}{x - x ^{3}} d x

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

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

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

Add a constant to the solution

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

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