What is the integral of (5x^2-3)/(x(x^2-1)) ?

The integral of (5x^2-3)/(x(x^2-1)) is 5/2 ln|x^2-1|-3(-ln|x|+1/2 ln|x+1|+1/2 ln|x-1|)+C

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integral of (5x^2-3)/(x(x^2-1))

\int \frac{5 x ^{2} - 3}{x ( x ^{2} - 1 )} d x

\frac{5}{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{5 x ^{2} - 3}{x ( x ^{2} - 1 )} d x

Expand

\frac{5 x ^{2} - 3}{x ( x ^{2} - 1 )} : \frac{5 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{5 x ^{2}}{x ( x ^{2} - 1 )} d x - \int \frac{3}{x ( x ^{2} - 1 )} d x

\int \frac{5 x ^{2}}{x ( x ^{2} - 1 )} d x = \frac{5}{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{5}{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{5}{2} ln x^{2} - 1 - 3 (- ln ∣ x ∣ + \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x - 1 ∣) + C

f (x) = \frac{4 x ^{2} + 8 x + 8}{x}

\frac{\partial}{\partial y} (x e^{x})

\int (tan^{2} (x) + tan^{4} (x)) d x

n = 2 \sum \infty \frac{2}{3 ^{n}}

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

What is the integral of (5x^2-3)/(x(x^2-1)) ?
The integral of (5x^2-3)/(x(x^2-1)) is 5/2 ln|x^2-1|-3(-ln|x|+1/2 ln|x+1|+1/2 ln|x-1|)+C