What is the integral of (1+x^2)/(x(x^2-1)) ?

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

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integral of (1+x^2)/(x(x^2-1))

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

- ln ∣ x ∣ + ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

Solution steps

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

Take the partial fraction of

\frac{1 + x ^{2}}{x ( x ^{2} - 1 )} : - \frac{1}{x} + \frac{1}{x + 1} + \frac{1}{x - 1}

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

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

= - ln ∣ x ∣ + ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣

Add a constant to the solution

= - ln ∣ x ∣ + ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

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