What is the integral of (x^2)/(sqrt(8-x)) ?

The integral of (x^2)/(sqrt(8-x)) is -2(1/5 (8-x)^{5/2}-16/3 (8-x)^{3/2}+64sqrt(8-x))+C

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integral of (x^2)/(sqrt(8-x))

\int \frac{x ^{2}}{8 - x} d x

- 2 (\frac{1}{5} (8 - x)^{\frac{5}{2}} - \frac{16}{3} (8 - x)^{\frac{3}{2}} + 64 8 - x) + C

Solution steps

\int \frac{x ^{2}}{8 - x} d x

Apply u-substitution

: \int - 2 (- u^{2} + 8)^{2} d u

= \int - 2 (- u^{2} + 8)^{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} + 8)^{2} d u

Expand

(- u^{2} + 8)^{2} : u^{4} - 16 u^{2} + 64

= - 2 \cdot \int u^{4} - 16 u^{2} + 64 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 16 u^{2} d u + \int 64 d u)

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

\int 16 u^{2} d u = \frac{16 u ^{3}}{3}

\int 64 d u = 64 u

= - 2 (\frac{u ^{5}}{5} - \frac{16 u ^{3}}{3} + 64 u)

Substitute back

u = 8 - x

= - 2 (\frac{( 8 - x ) ^{5}}{5} - \frac{16 ( 8 - x ) ^{3}}{3} + 64 8 - x)

Simplify

- 2 (\frac{( 8 - x ) ^{5}}{5} - \frac{16 ( 8 - x ) ^{3}}{3} + 64 8 - x) : - 2 (\frac{1}{5} (8 - x)^{\frac{5}{2}} - \frac{16}{3} (8 - x)^{\frac{3}{2}} + 64 8 - x)

= - 2 (\frac{1}{5} (8 - x)^{\frac{5}{2}} - \frac{16}{3} (8 - x)^{\frac{3}{2}} + 64 8 - x)

Add a constant to the solution

= - 2 (\frac{1}{5} (8 - x)^{\frac{5}{2}} - \frac{16}{3} (8 - x)^{\frac{3}{2}} + 64 8 - x) + C

(x^{2} + 15)^{2}

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

\frac{d}{d x} (\frac{1}{4} cos^{4} (x^{3}) - sin^{2} (6 x))

\int x sinh (2 x) d x

15 e^{x}

What is the integral of (x^2)/(sqrt(8-x)) ?
The integral of (x^2)/(sqrt(8-x)) is -2(1/5 (8-x)^{5/2}-16/3 (8-x)^{3/2}+64sqrt(8-x))+C