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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

x \to 5 - lim (\frac{x}{x - 5})

y^{^{''}} - 2 y^{^{'}} = 6

h \to 2 lim (∣ h - 2 ∣)

x \to 0 + lim ((tan (x))^{x})

2 t^{4}

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