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

The integral of (x^2)/(sqrt(1-x)) is -2(1/5 (1-x)^{5/2}-2/3 (1-x)^{3/2}+sqrt(1-x))+C

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integral of (x^2)/(sqrt(1-x))

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

- 2 (\frac{1}{5} (1 - x)^{\frac{5}{2}} - \frac{2}{3} (1 - x)^{\frac{3}{2}} + 1 - x) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

(- u^{2} + 1)^{2} : u^{4} - 2 u^{2} + 1

= - 2 \cdot \int u^{4} - 2 u^{2} + 1 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

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

\int u^{4} d u = \frac{u ^{5}}{5}

\int 2 u^{2} d u = \frac{2 u ^{3}}{3}

\int 1 d u = u

= - 2 (\frac{u ^{5}}{5} - \frac{2 u ^{3}}{3} + u)

Substitute back

u = 1 - x

= - 2 (\frac{( 1 - x ) ^{5}}{5} - \frac{2 ( 1 - x ) ^{3}}{3} + 1 - x)

Simplify

- 2 (\frac{( 1 - x ) ^{5}}{5} - \frac{2 ( 1 - x ) ^{3}}{3} + 1 - x) : - 2 (\frac{1}{5} (1 - x)^{\frac{5}{2}} - \frac{2}{3} (1 - x)^{\frac{3}{2}} + 1 - x)

= - 2 (\frac{1}{5} (1 - x)^{\frac{5}{2}} - \frac{2}{3} (1 - x)^{\frac{3}{2}} + 1 - x)

Add a constant to the solution

= - 2 (\frac{1}{5} (1 - x)^{\frac{5}{2}} - \frac{2}{3} (1 - x)^{\frac{3}{2}} + 1 - x) + C

θ^{^{''}} + \frac{4 k}{3} θ = 0

\frac{d}{d x} (\frac{e ^{x} ( x ^{2} + 2 x ) ^{3}}{3 x ^{2} + x})

y = \frac{4}{x ^{2} + 4}, [- 10, 10]

\frac{d}{d x} ((x - y)^{2})

4 x - 17

What is the integral of (x^2)/(sqrt(1-x)) ?
The integral of (x^2)/(sqrt(1-x)) is -2(1/5 (1-x)^{5/2}-2/3 (1-x)^{3/2}+sqrt(1-x))+C