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

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

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

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

- ln ∣ x ∣ + ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

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

= \int - \frac{1}{x} + \frac{1}{x - 1} d 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{1}{x} d x + \int \frac{1}{x - 1} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{1}{x - 1} d x = ln ∣ x - 1 ∣

= - ln ∣ x ∣ + ln ∣ x - 1 ∣

Add a constant to the solution

= - ln ∣ x ∣ + ln ∣ x - 1 ∣ + C

\frac{\partial}{\partial L} (K (α) (α)^{α} L^{1 - α})

y^{^{'}} + 4 y = 0

n = 1 \sum \infty 8 (\frac{7}{8})^{n}

\int 18 x^{2} (6 x^{3} + 5)^{\frac{1}{2}} d x

\int tan^{2} (x) (sec^{2} (x) - 1) d x

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