What is the integral of (x^2)/(sqrt(9-x)) ?

The integral of (x^2)/(sqrt(9-x)) is -2(1/5 (9-x)^{5/2}-6(9-x)^{3/2}+81sqrt(9-x))+C

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integral of (x^2)/(sqrt(9-x))

\int \frac{x ^{2}}{9 - x} d x

- 2 (\frac{1}{5} (9 - x)^{\frac{5}{2}} - 6 (9 - x)^{\frac{3}{2}} + 81 9 - x) + C

Solution steps

\int \frac{x ^{2}}{9 - x} d x

Apply u-substitution

: \int - 2 (- u^{2} + 9)^{2} d u

= \int - 2 (- u^{2} + 9)^{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} + 9)^{2} d u

Expand

(- u^{2} + 9)^{2} : u^{4} - 18 u^{2} + 81

= - 2 \cdot \int u^{4} - 18 u^{2} + 81 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 18 u^{2} d u + \int 81 d u)

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

\int 18 u^{2} d u = 6 u^{3}

\int 81 d u = 81 u

= - 2 (\frac{u ^{5}}{5} - 6 u^{3} + 81 u)

Substitute back

u = 9 - x

= - 2 (\frac{( 9 - x ) ^{5}}{5} - 6 (9 - x)^{3} + 81 9 - x)

Simplify

- 2 (\frac{( 9 - x ) ^{5}}{5} - 6 (9 - x)^{3} + 81 9 - x) : - 2 (\frac{1}{5} (9 - x)^{\frac{5}{2}} - 6 (9 - x)^{\frac{3}{2}} + 81 9 - x)

= - 2 (\frac{1}{5} (9 - x)^{\frac{5}{2}} - 6 (9 - x)^{\frac{3}{2}} + 81 9 - x)

Add a constant to the solution

= - 2 (\frac{1}{5} (9 - x)^{\frac{5}{2}} - 6 (9 - x)^{\frac{3}{2}} + 81 9 - x) + C

\frac{5 ( 1 - sin ( x ) )}{2 cos ( x )}

x^{^{'}} = x^{3} (t)

x^{3} + 2 x + 1

f (x) = \frac{5}{x ^{2}}

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

What is the integral of (x^2)/(sqrt(9-x)) ?
The integral of (x^2)/(sqrt(9-x)) is -2(1/5 (9-x)^{5/2}-6(9-x)^{3/2}+81sqrt(9-x))+C