What is the integral of (x^2-3)/(x(x^2-1)) ?

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

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integral of (x^2-3)/(x(x^2-1))

\int \frac{x ^{2} - 3}{x ( x ^{2} - 1 )} d x

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

Solution steps

\int \frac{x ^{2} - 3}{x ( x ^{2} - 1 )} d x

Expand

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

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

Add a constant to the solution

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

\int_{0}^{π} 4 sec^{2} (\frac{x}{4}) d x

\frac{\partial}{\partial x} (2 x^{3} cos (x y))

x \to 0 lim (\frac{1}{3 ^{x}})

f (x) = x^{3} + 5 x + 7

\frac{4}{s ^{2} + 2 s + 4}

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