How do we integrate the following without substitution!:
$$\int x\sqrt{1-x}\,dx$$
My try:
$$\int x\sqrt{1-x}\,dx=\int\dfrac{x(1-x)}{\sqrt{1-x}}\,dx=\int\dfrac{x}{\sqrt{1-x}}\,dx-\int\dfrac{x^2}{\sqrt{1-x}}\,dx=I_1-I_2$$
$$I_1=-\int\dfrac{1-x-1}{\sqrt{1-x}}\,dx=\\-\int\sqrt{1-x}\,dx-\int\dfrac{1}{\sqrt{1-x}}\,dx=\dfrac23(1-x)^\frac32-2(1-x)^\frac12+C_1$$
But I really don't see what to do with $I_2$, so maybe that's not the right approach.
$$\int x\sqrt{1-x}dx\\ =\int(1+x-1)\sqrt{1-x}dx \\=\int\sqrt{1-x}dx-\int(1-x)^{\frac32}dx\\ =-\frac23(1-x)^{\frac32}+\frac25(1-x)^{\frac52}+C $$ I took the idea of Math Student.