What is the integral of ((y^2-14y+5))/(y(y-1)(y-5)) ?

The integral of ((y^2-14y+5))/(y(y-1)(y-5)) is

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integral of ((y^2-14y+5))/(y(y-1)(y-5))

\int \frac{( y ^{2} - 14 y + 5 )}{y ( y - 1 ) ( y - 5 )} d y

\frac{1}{5} ln ∣ y ∣ + \frac{4}{5} ln ∣ y - 5 ∣ - 12 (- \frac{1}{5} ln ∣ y ∣ + \frac{1}{5} ln ∣ y - 5 ∣) - 8 (\frac{1}{5} ln ∣ y ∣ - \frac{1}{4} ln ∣ y - 1 ∣ + \frac{1}{20} ln ∣ y - 5 ∣) + C

Solution steps

\int \frac{y ^{2} - 14 y + 5}{y ( y - 1 ) ( y - 5 )} d y

Apply u-substitution

: \int \frac{u ^{2} - 12 u - 8}{u ( u + 1 ) ( u - 4 )} d u

= \int \frac{u ^{2} - 12 u - 8}{u ( u + 1 ) ( u - 4 )} d u

Expand

\frac{u ^{2} - 12 u - 8}{u ( u + 1 ) ( u - 4 )} : \frac{u ^{2}}{u ( u + 1 ) ( u - 4 )} - \frac{12 u}{u ( u + 1 ) ( u - 4 )} - \frac{8}{u ( u + 1 ) ( u - 4 )}

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{u ^{2}}{u ( u + 1 ) ( u - 4 )} d u - \int \frac{12 u}{u ( u + 1 ) ( u - 4 )} d u - \int \frac{8}{u ( u + 1 ) ( u - 4 )} d u

\int \frac{u ^{2}}{u ( u + 1 ) ( u - 4 )} d u = \frac{1}{5} ln ∣ u + 1 ∣ + \frac{4}{5} ln ∣ u - 4 ∣

\int \frac{12 u}{u ( u + 1 ) ( u - 4 )} d u = 12 (- \frac{1}{5} ln ∣ u + 1 ∣ + \frac{1}{5} ln ∣ u - 4 ∣)

\int \frac{8}{u ( u + 1 ) ( u - 4 )} d u = 8 (\frac{1}{5} ln ∣ u + 1 ∣ - \frac{1}{4} ln ∣ u ∣ + \frac{1}{20} ln ∣ u - 4 ∣)

= \frac{1}{5} ln ∣ u + 1 ∣ + \frac{4}{5} ln ∣ u - 4 ∣ - 12 (- \frac{1}{5} ln ∣ u + 1 ∣ + \frac{1}{5} ln ∣ u - 4 ∣) - 8 (\frac{1}{5} ln ∣ u + 1 ∣ - \frac{1}{4} ln ∣ u ∣ + \frac{1}{20} ln ∣ u - 4 ∣)

Substitute back

u = y - 1

= \frac{1}{5} ln ∣ y - 1 + 1 ∣ + \frac{4}{5} ln ∣ y - 1 - 4 ∣ - 12 (- \frac{1}{5} ln ∣ y - 1 + 1 ∣ + \frac{1}{5} ln ∣ y - 1 - 4 ∣) - 8 (\frac{1}{5} ln ∣ y - 1 + 1 ∣ - \frac{1}{4} ln ∣ y - 1 ∣ + \frac{1}{20} ln ∣ y - 1 - 4 ∣)

Simplify

\frac{1}{5} ln ∣ y - 1 + 1 ∣ + \frac{4}{5} ln ∣ y - 1 - 4 ∣ - 12 (- \frac{1}{5} ln ∣ y - 1 + 1 ∣ + \frac{1}{5} ln ∣ y - 1 - 4 ∣) - 8 (\frac{1}{5} ln ∣ y - 1 + 1 ∣ - \frac{1}{4} ln ∣ y - 1 ∣ + \frac{1}{20} ln ∣ y - 1 - 4 ∣) : \frac{1}{5} ln ∣ y ∣ + \frac{4}{5} ln ∣ y - 5 ∣ - 12 (- \frac{1}{5} ln ∣ y ∣ + \frac{1}{5} ln ∣ y - 5 ∣) - 8 (\frac{1}{5} ln ∣ y ∣ - \frac{1}{4} ln ∣ y - 1 ∣ + \frac{1}{20} ln ∣ y - 5 ∣)

= \frac{1}{5} ln ∣ y ∣ + \frac{4}{5} ln ∣ y - 5 ∣ - 12 (- \frac{1}{5} ln ∣ y ∣ + \frac{1}{5} ln ∣ y - 5 ∣) - 8 (\frac{1}{5} ln ∣ y ∣ - \frac{1}{4} ln ∣ y - 1 ∣ + \frac{1}{20} ln ∣ y - 5 ∣)

Add a constant to the solution

= \frac{1}{5} ln ∣ y ∣ + \frac{4}{5} ln ∣ y - 5 ∣ - 12 (- \frac{1}{5} ln ∣ y ∣ + \frac{1}{5} ln ∣ y - 5 ∣) - 8 (\frac{1}{5} ln ∣ y ∣ - \frac{1}{4} ln ∣ y - 1 ∣ + \frac{1}{20} ln ∣ y - 5 ∣) + C

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What is the integral of ((y^2-14y+5))/(y(y-1)(y-5)) ?
The integral of ((y^2-14y+5))/(y(y-1)(y-5)) is