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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

= 5 (ln ∣ y ∣ - ln ∣ y - 1 ∣ - \frac{1}{y - 1}) - (- \frac{2}{y - 1})

Simplify

5 (ln ∣ y ∣ - ln ∣ y - 1 ∣ - \frac{1}{y - 1}) - (- \frac{2}{y - 1}) : 5 ln ∣ y ∣ - 5 ln ∣ y - 1 ∣ - \frac{3}{y - 1}

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

Add a constant to the solution

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

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