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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

\frac{d y}{d x} = y (x y^{7} - 1)

\int sin^{3} (x co) s^{9} x d x

y = - 6 x^{2} + 5 x, (0, 0)

\frac{d y}{d t} + y = cos (t), y (0) = 1

\frac{1}{2} x

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