How do I integrate this? How do I integrate $\displaystyle\int (x^2 + 2)\sqrt{1-x} \; dx$ ?
I have feeling substitution might be used, but I just can't put my finger on it...
Thank you.

Note that $$x^2+2=(1-x)^2-2(1-x)+3,$$ which implies that
$$(x^2+2)\sqrt{1-x}=(1-x)^\frac{5}{2}-2(1-x)^{\frac{3}{2}}+3(1-x)^{\frac{1}{2}}.$$
Therefore, let $u=1-x$, we have $dx=-du$, which implies that
$$\int (x^2+2)\sqrt{1-x}dx=-\int u^\frac{5}{2}-2u^{\frac{3}{2}}+3u^{\frac{1}{2}}du$$
$$=-\frac{2}{7}u^{\frac{7}{2}}+\frac{4}{5}u^{\frac{5}{2}}+2u^{\frac{3}{2}}+C$$
$$=-\frac{2}{7}(1-x)^{\frac{7}{2}}+\frac{4}{5}(1-x)^{\frac{5}{2}}+2(1-x)^{\frac{3}{2}}+C.$$
