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

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

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integral of (9(y-1))/(y^2(y^2-9))

\int \frac{9 ( y - 1 )}{y ^{2} ( y ^{2} - 9 )} d y

- ln ∣ y ∣ - \frac{1}{y} + \frac{2}{3} ln ∣ y + 3 ∣ + \frac{1}{3} ln ∣ y - 3 ∣ + C

Solution steps

\int \frac{9 ( y - 1 )}{y ^{2} ( y ^{2} - 9 )} d y

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int \frac{y - 1}{y ^{2} ( y ^{2} - 9 )} d y

Take the partial fraction of

\frac{y - 1}{y ^{2} ( y ^{2} - 9 )} : - \frac{1}{9 y} + \frac{1}{9 y ^{2}} + \frac{2}{27 ( y + 3 )} + \frac{1}{27 ( y - 3 )}

= 9 \cdot \int - \frac{1}{9 y} + \frac{1}{9 y ^{2}} + \frac{2}{27 ( y + 3 )} + \frac{1}{27 ( y - 3 )} d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 9 (- \int \frac{1}{9 y} d y + \int \frac{1}{9 y ^{2}} d y + \int \frac{2}{27 ( y + 3 )} d y + \int \frac{1}{27 ( y - 3 )} d y)

\int \frac{1}{9 y} d y = \frac{1}{9} ln ∣ y ∣

\int \frac{1}{9 y ^{2}} d y = - \frac{1}{9 y}

\int \frac{2}{27 ( y + 3 )} d y = \frac{2}{27} ln ∣ y + 3 ∣

\int \frac{1}{27 ( y - 3 )} d y = \frac{1}{27} ln ∣ y - 3 ∣

= 9 (- \frac{1}{9} ln ∣ y ∣ - \frac{1}{9 y} + \frac{2}{27} ln ∣ y + 3 ∣ + \frac{1}{27} ln ∣ y - 3 ∣)

Simplify

9 (- \frac{1}{9} ln ∣ y ∣ - \frac{1}{9 y} + \frac{2}{27} ln ∣ y + 3 ∣ + \frac{1}{27} ln ∣ y - 3 ∣) : - ln ∣ y ∣ - \frac{1}{y} + \frac{2}{3} ln ∣ y + 3 ∣ + \frac{1}{3} ln ∣ y - 3 ∣

= - ln ∣ y ∣ - \frac{1}{y} + \frac{2}{3} ln ∣ y + 3 ∣ + \frac{1}{3} ln ∣ y - 3 ∣

Add a constant to the solution

= - ln ∣ y ∣ - \frac{1}{y} + \frac{2}{3} ln ∣ y + 3 ∣ + \frac{1}{3} ln ∣ y - 3 ∣ + C

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