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

The integral of (3t^2+3)/(t-1) is 3/2 (t-1)^2+6t-6+6ln|t-1|+C

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

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

\frac{3}{2} (t - 1)^{2} + 6 t - 6 + 6 ln ∣ t - 1 ∣ + C

Solution steps

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

Expand

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

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

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

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

Simplify

3 (\frac{( t - 1 ) ^{2}}{2} + 2 (t - 1) + ln ∣ t - 1 ∣) + 3 ln ∣ t - 1 ∣ : \frac{3}{2} (t - 1)^{2} + 6 t - 6 + 6 ln ∣ t - 1 ∣

= \frac{3}{2} (t - 1)^{2} + 6 t - 6 + 6 ln ∣ t - 1 ∣

Add a constant to the solution

= \frac{3}{2} (t - 1)^{2} + 6 t - 6 + 6 ln ∣ t - 1 ∣ + C

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What is the integral of (3t^2+3)/(t-1) ?
The integral of (3t^2+3)/(t-1) is 3/2 (t-1)^2+6t-6+6ln|t-1|+C