I don't know how to solve indefinite integral of [x^2 * sqrt(3-x)].

Question

anonymous · Answer

$$\int\limits x^2 \left ( -x+3 \right )^{\frac{1}{2}} dx$$

dumbcow · Answer

substitution
u = 3-x   -->  x = 3-u
du = -dx

$$= -\int\limits (3-u)^{2} \sqrt{u} du$$
$$= -\int\limits \sqrt{u}(u^{2}-6u+9) du$$
distribute and then integrate each term using power rule

anonymous · Answer

thank you

dumbcow · Answer

yw

anonymous · Answer

$$\int\limits x^2 \sqrt{3-x}\:dx = -\frac{2}{7}(3-x)^\frac{7}{2} + \frac{12}{5}(3-x)^\frac{5}{2} - 6(3-x)^\frac{3}{2} + C$$

dumbcow · Answer

the coefficients seem to be a little off http://www.wolframalpha.com/input/?i=integral+x%5E2+sqrt%283-x%29+dx

anonymous · Answer

Yet http://www.wolframalpha.com/input/?i=derivative+of+%28-2%2F7%29%283-x%29^%287%2F2%29%2B%2812%2F5%29%283-x%29^%285%2F2%29-6%283-x%29^%283%2F2%29