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

The integral of (x^2)/(sqrt(2-x)) is -2(1/5 (2-x)^{5/2}-4/3 (2-x)^{3/2}+4sqrt(2-x))+C

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

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

- 2 (\frac{1}{5} (2 - x)^{\frac{5}{2}} - \frac{4}{3} (2 - x)^{\frac{3}{2}} + 4 2 - x) + C

Solution steps

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

Apply u-substitution

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

Expand

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

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

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

\int 4 u^{2} d u = \frac{4 u ^{3}}{3}

\int 4 d u = 4 u

= - 2 (\frac{u ^{5}}{5} - \frac{4 u ^{3}}{3} + 4 u)

Substitute back

u = 2 - x

= - 2 (\frac{( 2 - x ) ^{5}}{5} - \frac{4 ( 2 - x ) ^{3}}{3} + 4 2 - x)

Simplify

- 2 (\frac{( 2 - x ) ^{5}}{5} - \frac{4 ( 2 - x ) ^{3}}{3} + 4 2 - x) : - 2 (\frac{1}{5} (2 - x)^{\frac{5}{2}} - \frac{4}{3} (2 - x)^{\frac{3}{2}} + 4 2 - x)

= - 2 (\frac{1}{5} (2 - x)^{\frac{5}{2}} - \frac{4}{3} (2 - x)^{\frac{3}{2}} + 4 2 - x)

Add a constant to the solution

= - 2 (\frac{1}{5} (2 - x)^{\frac{5}{2}} - \frac{4}{3} (2 - x)^{\frac{3}{2}} + 4 2 - x) + C

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What is the integral of (x^2)/(sqrt(2-x)) ?
The integral of (x^2)/(sqrt(2-x)) is -2(1/5 (2-x)^{5/2}-4/3 (2-x)^{3/2}+4sqrt(2-x))+C