What is the integral of (3+u)/(u^2-1) ?

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

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integral of (3+u)/(u^2-1)

\int \frac{3 + u}{u ^{2} - 1} d u

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

Solution steps

\int \frac{3 + u}{u ^{2} - 1} d u

Expand

\frac{3 + u}{u ^{2} - 1} : \frac{3}{u ^{2} - 1} + \frac{u}{u ^{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{3}{u ^{2} - 1} d u + \int \frac{u}{u ^{2} - 1} d u

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

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

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

Add a constant to the solution

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

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