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

The integral of (y^2-2y+4)/(y(y-2)^2) is -2/(y-2)+ln|y|+C

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integral of (y^2-2y+4)/(y(y-2)^2)

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

- \frac{2}{y - 2} + ln ∣ y ∣ + C

Solution steps

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

Expand

\frac{y ^{2} - 2 y + 4}{y ( y - 2 ) ^{2}} : \frac{y ^{2}}{y ( y - 2 ) ^{2}} - \frac{2 y}{y ( y - 2 ) ^{2}} + \frac{4}{y ( y - 2 ) ^{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{y ^{2}}{y ( y - 2 ) ^{2}} d y - \int \frac{2 y}{y ( y - 2 ) ^{2}} d y + \int \frac{4}{y ( y - 2 ) ^{2}} d y

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

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

\int \frac{4}{y ( y - 2 ) ^{2}} d y = ln ∣ y ∣ - ln ∣ y - 2 ∣ - \frac{2}{y - 2}

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

Simplify

ln ∣ y - 2 ∣ - \frac{2}{y - 2} - (- \frac{2}{y - 2}) + ln ∣ y ∣ - ln ∣ y - 2 ∣ - \frac{2}{y - 2} : - \frac{2}{y - 2} + ln ∣ y ∣

= - \frac{2}{y - 2} + ln ∣ y ∣

Add a constant to the solution

= - \frac{2}{y - 2} + ln ∣ y ∣ + C

y^{^{''''}} - y^{^{''}} = 4 x + 2 x e^{- x}

s l o p e (1.4) (7.1)

\int_{3}^{9} \frac{1}{x ( ln ( x ) ) ^{2}} d x

d er i v a t i v e f (x) = \frac{1 - 4 x}{3 + x}

x \to 5 lim (ln (x - 4) + \frac{1}{2})

What is the integral of (y^2-2y+4)/(y(y-2)^2) ?
The integral of (y^2-2y+4)/(y(y-2)^2) is -2/(y-2)+ln|y|+C