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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \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 = 1 - y

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

Simplify

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

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

Add a constant to the solution

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

f (x) = 3 x - 2 x^{- \frac{1}{2}} + 1, at x = 1

10 y^{^{''}} + 1000 y = 20 cos (8 t), y (0) = 0, y^{^{'}} (0) = 0

y = 3 x - 2

y^{^{''}} + 6 y^{^{'}} + 34 y = 0

x \to 0 lim (\frac{5 x ^{3}}{x})

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