What is the integral of (x^2)/(sqrt(6-x)) ?

The integral of (x^2)/(sqrt(6-x)) is -2(1/5 (6-x)^{5/2}-4(6-x)^{3/2}+36sqrt(6-x))+C

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integral of (x^2)/(sqrt(6-x))

\int \frac{x ^{2}}{6 - x} d x

- 2 (\frac{1}{5} (6 - x)^{\frac{5}{2}} - 4 (6 - x)^{\frac{3}{2}} + 36 6 - x) + C

Solution steps

\int \frac{x ^{2}}{6 - x} d x

Apply u-substitution

: \int - 2 (- u^{2} + 6)^{2} d u

= \int - 2 (- u^{2} + 6)^{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} + 6)^{2} d u

Expand

(- u^{2} + 6)^{2} : u^{4} - 12 u^{2} + 36

= - 2 \cdot \int u^{4} - 12 u^{2} + 36 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 12 u^{2} d u + \int 36 d u)

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

\int 12 u^{2} d u = 4 u^{3}

\int 36 d u = 36 u

= - 2 (\frac{u ^{5}}{5} - 4 u^{3} + 36 u)

Substitute back

u = 6 - x

= - 2 (\frac{( 6 - x ) ^{5}}{5} - 4 (6 - x)^{3} + 36 6 - x)

Simplify

- 2 (\frac{( 6 - x ) ^{5}}{5} - 4 (6 - x)^{3} + 36 6 - x) : - 2 (\frac{1}{5} (6 - x)^{\frac{5}{2}} - 4 (6 - x)^{\frac{3}{2}} + 36 6 - x)

= - 2 (\frac{1}{5} (6 - x)^{\frac{5}{2}} - 4 (6 - x)^{\frac{3}{2}} + 36 6 - x)

Add a constant to the solution

= - 2 (\frac{1}{5} (6 - x)^{\frac{5}{2}} - 4 (6 - x)^{\frac{3}{2}} + 36 6 - x) + C

x \to 2 lim (f (x))

y = x^{3} - 9 x^{2} + 18 x, y = - x^{3} + 9 x^{2} - 18 x

\int x sin (6 - x) d x

\int \frac{16}{x ^{2} + 1} d x

\int \frac{- cos ( x )}{sin ^{8} ( x )} d x

What is the integral of (x^2)/(sqrt(6-x)) ?
The integral of (x^2)/(sqrt(6-x)) is -2(1/5 (6-x)^{5/2}-4(6-x)^{3/2}+36sqrt(6-x))+C