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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

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