Let, I=intx/sqrt(1-x)dx=-int(-x)/sqrt(1-x)dxI=∫x√1−xdx=−∫−x√1−xdx,
=-int{(1-x)-1}/sqrt(1-x)dx=−∫(1−x)−1√1−xdx,
=-int{(1-x)/sqrt(1-x)-1/sqrt(1-x)}dx=−∫{1−x√1−x−1√1−x}dx,
=-int{sqrt(1-x)-1/sqrt(1-x)}dx=−∫{√1−x−1√1−x}dx,
=-int(1-x)^(1/2)dx+int(1-x)^(-1/2)dx=−∫(1−x)12dx+∫(1−x)−12dx.
Recall that,
intf(x)dx=F(x)+C rArr intf(ax+b)dx=1/aF(ax+b)+K, (a!=0)∫f(x)dx=F(x)+C⇒∫f(ax+b)dx=1aF(ax+b)+K,(a≠0)
E.g, intx^(1/2)dx=2/3x^(3/2)+C:.int(2-3x)^(1/2)dx=1/(-3)(2-3x)^(3/2)+K.
:. I=-1/(-1)(1-x)^(1/2+1)/(1/2+1)+1/(-1)(1-x)^(-1/2+1)/(-1/2+1),
=2/3(1-x)^(3/2)-2(1-x)^(1/2),
=2/3(1-x)^(1/2){(1-x)-3}.
rArr I=-2/3(2+x)sqrt(1-x)+C.