What is the integral of (y^2+3y-1)/(y^3+6y^2+9y) ?

The integral of (y^2+3y-1)/(y^3+6y^2+9y) is -1/9 ln|y|+10/9 ln|y+3|-1/(3(y+3))+C

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integral of (y^2+3y-1)/(y^3+6y^2+9y)

\int \frac{y ^{2} + 3 y - 1}{y ^{3} + 6 y ^{2} + 9 y} d y

- \frac{1}{9} ln ∣ y ∣ + \frac{10}{9} ln ∣ y + 3 ∣ - \frac{1}{3 ( y + 3 )} + C

Solution steps

\int \frac{y ^{2} + 3 y - 1}{y ^{3} + 6 y ^{2} + 9 y} d y

Expand

\frac{y ^{2} + 3 y - 1}{y ^{3} + 6 y ^{2} + 9 y} : \frac{y ^{2}}{y ^{3} + 6 y ^{2} + 9 y} + \frac{3 y}{y ^{3} + 6 y ^{2} + 9 y} - \frac{1}{y ^{3} + 6 y ^{2} + 9 y}

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 ^{2}}{y ^{3} + 6 y ^{2} + 9 y} d y + \int \frac{3 y}{y ^{3} + 6 y ^{2} + 9 y} d y - \int \frac{1}{y ^{3} + 6 y ^{2} + 9 y} d y

\int \frac{y ^{2}}{y ^{3} + 6 y ^{2} + 9 y} d y = ln ∣ y + 3 ∣ + \frac{3}{y + 3}

\int \frac{3 y}{y ^{3} + 6 y ^{2} + 9 y} d y = - \frac{3}{y + 3}

\int \frac{1}{y ^{3} + 6 y ^{2} + 9 y} d y = \frac{1}{9} ln ∣ y ∣ - \frac{1}{9} ln ∣ y + 3 ∣ + \frac{1}{3 ( y + 3 )}

= ln ∣ y + 3 ∣ + \frac{3}{y + 3} - \frac{3}{y + 3} - (\frac{1}{9} ln ∣ y ∣ - \frac{1}{9} ln ∣ y + 3 ∣ + \frac{1}{3 ( y + 3 )})

Simplify

ln ∣ y + 3 ∣ + \frac{3}{y + 3} - \frac{3}{y + 3} - (\frac{1}{9} ln ∣ y ∣ - \frac{1}{9} ln ∣ y + 3 ∣ + \frac{1}{3 ( y + 3 )}) : - \frac{1}{9} ln ∣ y ∣ + \frac{10}{9} ln ∣ y + 3 ∣ - \frac{1}{3 ( y + 3 )}

= - \frac{1}{9} ln ∣ y ∣ + \frac{10}{9} ln ∣ y + 3 ∣ - \frac{1}{3 ( y + 3 )}

Add a constant to the solution

= - \frac{1}{9} ln ∣ y ∣ + \frac{10}{9} ln ∣ y + 3 ∣ - \frac{1}{3 ( y + 3 )} + C

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What is the integral of (y^2+3y-1)/(y^3+6y^2+9y) ?
The integral of (y^2+3y-1)/(y^3+6y^2+9y) is -1/9 ln|y|+10/9 ln|y+3|-1/(3(y+3))+C