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

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

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

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

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

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Take the partial fraction of

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{t} d t = ln ∣ t ∣

\int \frac{1}{t ^{2}} d t = - \frac{1}{t}

\int \frac{1}{t - 1} d t = ln ∣ t - 1 ∣

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

Simplify

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

Add a constant to the solution

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

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