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

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

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

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

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

Solution steps

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

Expand

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

\int \frac{5 x}{x ^{2} - 1} d x = \frac{5}{2} ln x^{2} - 1

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

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

Simplify

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

y = \frac{6}{x}

\int 7 (sin (x))^{4} (cos (x))^{2} d x

x \frac{d y}{d x} = x^{3} + 14 x^{3} y

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

\int c x y + \frac{1}{4}

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