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

The integral of (y^3-y-1)/(y-2) is 3y^2-y+5ln|y-2|+1/3 (y-2)^3-10+C

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

\int \frac{y ^{3} - y - 1}{y - 2} d y

3 y^{2} - y + 5 ln ∣ y - 2 ∣ + \frac{1}{3} (y - 2)^{3} - 10 + C

Solution steps

\int \frac{y ^{3} - y - 1}{y - 2} d y

Expand

\frac{y ^{3} - y - 1}{y - 2} : \frac{y ^{3}}{y - 2} - \frac{y}{y - 2} - \frac{1}{y - 2}

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{y ^{3}}{y - 2} d y - \int \frac{y}{y - 2} d y - \int \frac{1}{y - 2} d y

\int \frac{y ^{3}}{y - 2} d y = 3 y^{2} + 8 ln ∣ y - 2 ∣ + \frac{1}{3} (y - 2)^{3} - 12

\int \frac{y}{y - 2} d y = y - 2 + 2 ln ∣ y - 2 ∣

\int \frac{1}{y - 2} d y = ln ∣ y - 2 ∣

= 3 y^{2} + 8 ln ∣ y - 2 ∣ + \frac{1}{3} (y - 2)^{3} - 12 - (y - 2 + 2 ln ∣ y - 2 ∣) - ln ∣ y - 2 ∣

Simplify

3 y^{2} + 8 ln ∣ y - 2 ∣ + \frac{1}{3} (y - 2)^{3} - 12 - (y - 2 + 2 ln ∣ y - 2 ∣) - ln ∣ y - 2 ∣ : 3 y^{2} - y + 5 ln ∣ y - 2 ∣ + \frac{1}{3} (y - 2)^{3} - 10

= 3 y^{2} - y + 5 ln ∣ y - 2 ∣ + \frac{1}{3} (y - 2)^{3} - 10

Add a constant to the solution

= 3 y^{2} - y + 5 ln ∣ y - 2 ∣ + \frac{1}{3} (y - 2)^{3} - 10 + C

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What is the integral of (y^3-y-1)/(y-2) ?
The integral of (y^3-y-1)/(y-2) is 3y^2-y+5ln|y-2|+1/3 (y-2)^3-10+C