What is the integral of (4y^2-7y-12)/(y(y+2)(y-3)) ?

The integral of (4y^2-7y-12)/(y(y+2)(y-3)) is 2ln|y|+9/5 ln|y+2|+1/5 ln|y-3|+C

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integral of (4y^2-7y-12)/(y(y+2)(y-3))

\int \frac{4 y ^{2} - 7 y - 12}{y ( y + 2 ) ( y - 3 )} d y

2 ln ∣ y ∣ + \frac{9}{5} ln ∣ y + 2 ∣ + \frac{1}{5} ln ∣ y - 3 ∣ + C

Solution steps

\int \frac{4 y ^{2} - 7 y - 12}{y ( y + 2 ) ( y - 3 )} d y

Take the partial fraction of

\frac{4 y ^{2} - 7 y - 12}{y ( y + 2 ) ( y - 3 )} : \frac{2}{y} + \frac{9}{5 ( y + 2 )} + \frac{1}{5 ( y - 3 )}

= \int \frac{2}{y} + \frac{9}{5 ( y + 2 )} + \frac{1}{5 ( 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

= \int \frac{2}{y} d y + \int \frac{9}{5 ( y + 2 )} d y + \int \frac{1}{5 ( y - 3 )} d y

\int \frac{2}{y} d y = 2 ln ∣ y ∣

\int \frac{9}{5 ( y + 2 )} d y = \frac{9}{5} ln ∣ y + 2 ∣

\int \frac{1}{5 ( y - 3 )} d y = \frac{1}{5} ln ∣ y - 3 ∣

= 2 ln ∣ y ∣ + \frac{9}{5} ln ∣ y + 2 ∣ + \frac{1}{5} ln ∣ y - 3 ∣

Add a constant to the solution

= 2 ln ∣ y ∣ + \frac{9}{5} ln ∣ y + 2 ∣ + \frac{1}{5} ln ∣ y - 3 ∣ + C

What is the integral of (4y^2-7y-12)/(y(y+2)(y-3)) ?
The integral of (4y^2-7y-12)/(y(y+2)(y-3)) is 2ln|y|+9/5 ln|y+2|+1/5 ln|y-3|+C