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

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

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

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

\frac{1}{2} ln x^{2} - 1 - ln ∣ x + 1 ∣ + ln ∣ x ∣ + C

Solution steps

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

Expand

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

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{2} ln x^{2} - 1 - ln ∣ x + 1 ∣ + ln ∣ x ∣ + C

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