Answer
$\frac{-1}{8}(1-x)^8+\frac{4}{7}(1-x)^7-\frac{2}{3}(1-x)^6$
Work Step by Step
Let u=1-x.
Then du=-1dx and (-1)du=dx
and x=1-u.
thus $\int (x+1)^2(1-x)^5dx$
=$\int (2-u)^2 u^5(-1)du$
=$\int (-u^7 +4u^6- 4u^5)du$
=$\frac{-1}{8}u^8+\frac{4}{7}u^7-\frac{2}{3}u^6$
=$\frac{-1}{8}(1-x)^8+\frac{4}{7}(1-x)^7-\frac{2}{3}(1-x)^6+C$
