What is the integral of (6z^2-5)/(z-1) ?

The integral of (6z^2-5)/(z-1) is 3z^2+6z+ln|z-1|-9+C

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integral of (6z^2-5)/(z-1)

\int \frac{6 z ^{2} - 5}{z - 1} d z

3 z^{2} + 6 z + ln ∣ z - 1 ∣ - 9 + C

Solution steps

\int \frac{6 z ^{2} - 5}{z - 1} d z

Expand

\frac{6 z ^{2} - 5}{z - 1} : \frac{6 z ^{2}}{z - 1} - \frac{5}{z - 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{6 z ^{2}}{z - 1} d z - \int \frac{5}{z - 1} d z

\int \frac{6 z ^{2}}{z - 1} d z = 6 (\frac{( z - 1 ) ^{2}}{2} + 2 (z - 1) + ln ∣ z - 1 ∣)

\int \frac{5}{z - 1} d z = 5 ln ∣ z - 1 ∣

= 6 (\frac{( z - 1 ) ^{2}}{2} + 2 (z - 1) + ln ∣ z - 1 ∣) - 5 ln ∣ z - 1 ∣

Simplify

6 (\frac{( z - 1 ) ^{2}}{2} + 2 (z - 1) + ln ∣ z - 1 ∣) - 5 ln ∣ z - 1 ∣ : 3 z^{2} + 6 z + ln ∣ z - 1 ∣ - 9

= 3 z^{2} + 6 z + ln ∣ z - 1 ∣ - 9

Add a constant to the solution

= 3 z^{2} + 6 z + ln ∣ z - 1 ∣ - 9 + C

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What is the integral of (6z^2-5)/(z-1) ?
The integral of (6z^2-5)/(z-1) is 3z^2+6z+ln|z-1|-9+C