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

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

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

\int \frac{3 t + 1}{t ^{2} ( t ^{2} - 1 )} d t

ln ∣ t + 1 ∣ + 2 ln ∣ t - 1 ∣ - 3 ln ∣ t ∣ + \frac{1}{t} + C

Solution steps

\int \frac{3 t + 1}{t ^{2} ( t ^{2} - 1 )} d t

Expand

\frac{3 t + 1}{t ^{2} ( t ^{2} - 1 )} : \frac{3 t}{t ^{2} ( t ^{2} - 1 )} + \frac{1}{t ^{2} ( t ^{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 t}{t ^{2} ( t ^{2} - 1 )} d t + \int \frac{1}{t ^{2} ( t ^{2} - 1 )} d t

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

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

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

Simplify

3 (- ln ∣ t ∣ + \frac{1}{2} ln ∣ t + 1 ∣ + \frac{1}{2} ln ∣ t - 1 ∣) + \frac{1}{t} - \frac{1}{2} ln ∣ t + 1 ∣ + \frac{1}{2} ln ∣ t - 1 ∣ : ln ∣ t + 1 ∣ + 2 ln ∣ t - 1 ∣ - 3 ln ∣ t ∣ + \frac{1}{t}

= ln ∣ t + 1 ∣ + 2 ln ∣ t - 1 ∣ - 3 ln ∣ t ∣ + \frac{1}{t}

Add a constant to the solution

= ln ∣ t + 1 ∣ + 2 ln ∣ t - 1 ∣ - 3 ln ∣ t ∣ + \frac{1}{t} + C

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