a)  Theo bài ra: $u=1-x => du=-dx$

=> $x=1-u$

$x=0=> u=1$

$x=1=>u=0$

=> $\int_{0}^{1}x(1-x)^{5}dx=-\int_{0}^{1}(1-u)u^{5}du=\frac{1}{42}$

b) Đặt $u=x => du=dx$

           $dv=(1-x)^{5}dx=> v=-\frac{(1-x)^{6}}{6}$

=> $\int_{0}^{1}x(1-x)^{5}dx=-\frac{x(1-x)^{6}}{6}\left |^{1}_{0}  \right |+\frac{1}{6}\int_{0}^{1}(1-x)^{6}dx=\frac{1}{42}$