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

The integral of (3t^2-3)/(t^3-3t) is

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

\int \frac{3 t ^{2} - 3}{t ^{3} - 3 t} d t

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

Solution steps

\int \frac{3 t ^{2} - 3}{t ^{3} - 3 t} d t

Expand

\frac{3 t ^{2} - 3}{t ^{3} - 3 t} : \frac{3 t ^{2}}{t ^{3} - 3 t} - \frac{3}{t ^{3} - 3 t}

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 ^{2}}{t ^{3} - 3 t} d t - \int \frac{3}{t ^{3} - 3 t} d t

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

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

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

Simplify

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

Add a constant to the solution

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

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