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

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

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

\int \frac{5 x + 1}{( x + 1 ) ( x - 1 )} d x

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

Solution steps

\int \frac{5 x + 1}{( x + 1 ) ( x - 1 )} d x

Expand

\frac{5 x + 1}{( x + 1 ) ( x - 1 )} : \frac{5 x}{( x + 1 ) ( x - 1 )} + \frac{1}{( x + 1 ) ( x - 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 + 1 ) ( x - 1 )} d x + \int \frac{1}{( x + 1 ) ( x - 1 )} d x

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

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

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

Add a constant to the solution

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

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