What is the integral of (x^2)/(sqrt(3-x)) ?

The integral of (x^2)/(sqrt(3-x)) is -2(1/5 (3-x)^{5/2}-2(3-x)^{3/2}+9sqrt(3-x))+C

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integral of (x^2)/(sqrt(3-x))

\int \frac{x ^{2}}{3 - x} d x

- 2 (\frac{1}{5} (3 - x)^{\frac{5}{2}} - 2 (3 - x)^{\frac{3}{2}} + 9 3 - x) + C

Solution steps

\int \frac{x ^{2}}{3 - x} d x

Apply u-substitution

: \int - 2 (- u^{2} + 3)^{2} d u

= \int - 2 (- u^{2} + 3)^{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} + 3)^{2} d u

Expand

(- u^{2} + 3)^{2} : u^{4} - 6 u^{2} + 9

= - 2 \cdot \int u^{4} - 6 u^{2} + 9 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 6 u^{2} d u + \int 9 d u)

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

\int 6 u^{2} d u = 2 u^{3}

\int 9 d u = 9 u

= - 2 (\frac{u ^{5}}{5} - 2 u^{3} + 9 u)

Substitute back

u = 3 - x

= - 2 (\frac{( 3 - x ) ^{5}}{5} - 2 (3 - x)^{3} + 9 3 - x)

Simplify

- 2 (\frac{( 3 - x ) ^{5}}{5} - 2 (3 - x)^{3} + 9 3 - x) : - 2 (\frac{1}{5} (3 - x)^{\frac{5}{2}} - 2 (3 - x)^{\frac{3}{2}} + 9 3 - x)

= - 2 (\frac{1}{5} (3 - x)^{\frac{5}{2}} - 2 (3 - x)^{\frac{3}{2}} + 9 3 - x)

Add a constant to the solution

= - 2 (\frac{1}{5} (3 - x)^{\frac{5}{2}} - 2 (3 - x)^{\frac{3}{2}} + 9 3 - x) + C

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What is the integral of (x^2)/(sqrt(3-x)) ?
The integral of (x^2)/(sqrt(3-x)) is -2(1/5 (3-x)^{5/2}-2(3-x)^{3/2}+9sqrt(3-x))+C