What is the integral of ((y-1)(y-5))/(y^3-y^2+3y+5) ?

The integral of ((y-1)(y-5))/(y^3-y^2+3y+5) is 3/2 ln|y+1|+1/4 (-ln|y^2-2y+5|-6arctan(1/2 (y-1)))+C

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integral of ((y-1)(y-5))/(y^3-y^2+3y+5)

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

\frac{3}{2} ln ∣ y + 1 ∣ + \frac{1}{4} (- ln y^{2} - 2 y + 5 - 6 arctan (\frac{1}{2} (y - 1))) + C

Solution steps

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

Take the partial fraction of

\frac{( y - 1 ) ( y - 5 )}{y ^{3} - y ^{2} + 3 y + 5} : \frac{3}{2 ( y + 1 )} + \frac{- y - 5}{2 ( y ^{2} - 2 y + 5 )}

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

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

\int \frac{- y - 5}{2 ( y ^{2} - 2 y + 5 )} d y = \frac{1}{4} (- ln y^{2} - 2 y + 5 - 6 arctan (\frac{1}{2} (y - 1)))

= \frac{3}{2} ln ∣ y + 1 ∣ + \frac{1}{4} (- ln y^{2} - 2 y + 5 - 6 arctan (\frac{1}{2} (y - 1)))

Add a constant to the solution

= \frac{3}{2} ln ∣ y + 1 ∣ + \frac{1}{4} (- ln y^{2} - 2 y + 5 - 6 arctan (\frac{1}{2} (y - 1))) + C

What is the integral of ((y-1)(y-5))/(y^3-y^2+3y+5) ?
The integral of ((y-1)(y-5))/(y^3-y^2+3y+5) is 3/2 ln|y+1|+1/4 (-ln|y^2-2y+5|-6arctan(1/2 (y-1)))+C