What is the integral of (x^2)/(sqrt(7-x)) ?

The integral of (x^2)/(sqrt(7-x)) is -2(1/5 (7-x)^{5/2}-14/3 (7-x)^{3/2}+49sqrt(7-x))+C

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integral of (x^2)/(sqrt(7-x))

\int \frac{x ^{2}}{7 - x} d x

- 2 (\frac{1}{5} (7 - x)^{\frac{5}{2}} - \frac{14}{3} (7 - x)^{\frac{3}{2}} + 49 7 - x) + C

Solution steps

\int \frac{x ^{2}}{7 - x} d x

Apply u-substitution

: \int - 2 (- u^{2} + 7)^{2} d u

= \int - 2 (- u^{2} + 7)^{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} + 7)^{2} d u

Expand

(- u^{2} + 7)^{2} : u^{4} - 14 u^{2} + 49

= - 2 \cdot \int u^{4} - 14 u^{2} + 49 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 14 u^{2} d u + \int 49 d u)

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

\int 14 u^{2} d u = \frac{14 u ^{3}}{3}

\int 49 d u = 49 u

= - 2 (\frac{u ^{5}}{5} - \frac{14 u ^{3}}{3} + 49 u)

Substitute back

u = 7 - x

= - 2 (\frac{( 7 - x ) ^{5}}{5} - \frac{14 ( 7 - x ) ^{3}}{3} + 49 7 - x)

Simplify

- 2 (\frac{( 7 - x ) ^{5}}{5} - \frac{14 ( 7 - x ) ^{3}}{3} + 49 7 - x) : - 2 (\frac{1}{5} (7 - x)^{\frac{5}{2}} - \frac{14}{3} (7 - x)^{\frac{3}{2}} + 49 7 - x)

= - 2 (\frac{1}{5} (7 - x)^{\frac{5}{2}} - \frac{14}{3} (7 - x)^{\frac{3}{2}} + 49 7 - x)

Add a constant to the solution

= - 2 (\frac{1}{5} (7 - x)^{\frac{5}{2}} - \frac{14}{3} (7 - x)^{\frac{3}{2}} + 49 7 - x) + C

n = 0 \sum \infty (\frac{x}{7})^{n}

(3, - 2), (- 5, 0)

x \to 0 lim (x \cdot 3^{\frac{1}{x}})

f (x) = tan^{5} (x)

\int \frac{( x ^{2} - 2 x + 1 )}{x ^{2} + x} d x

What is the integral of (x^2)/(sqrt(7-x)) ?
The integral of (x^2)/(sqrt(7-x)) is -2(1/5 (7-x)^{5/2}-14/3 (7-x)^{3/2}+49sqrt(7-x))+C