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

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

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

\int \frac{y ^{2}}{y - 1} d y

\frac{( y - 1 ) ^{2}}{2} + 2 (y - 1) + ln ∣ y - 1 ∣ + C

Solution steps

\int \frac{y ^{2}}{y - 1} d y

Apply u-substitution

= \int \frac{( u + 1 ) ^{2}}{u} d u

Expand

\frac{( u + 1 ) ^{2}}{u} : u + 2 + \frac{1}{u}

= \int u + 2 + \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u d u + \int 2 d u + \int \frac{1}{u} d u

\int u d u = \frac{u ^{2}}{2}

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{2}}{2} + 2 u + ln ∣ u ∣

Substitute back

u = y - 1

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

Add a constant to the solution

= \frac{( y - 1 ) ^{2}}{2} + 2 (y - 1) + ln ∣ y - 1 ∣ + C

\frac{d}{d x} ((A x + B) e^{- 2 x})

x \to 90 lim (tan (x))

\int \frac{e ^{x}}{x ^{4}} d x

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

\frac{d}{d x} (\frac{ln ( 3 x ^{2} - 6 x )}{4 x ^{3} - 2})

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