What is the integration of `(xdx)/sqrt(1-x)`??

Let, `u=sqrt(1-x)`

or, `u^2=1-x`

or, `x=1-u^2`

or, `dx=-2udu`

Now, `int (xdx)/(sqrt(1-x))=int (1-u^2)(-2udu)/u=int 2u^2du -int 2du`

Now, `int 2u^2 du -int 2du`

`=(2u^3)/3 - 2(u) +C=2/3u(u^2-3)+C=2/3sqrt(1-x){(1-x)-3} +C=2/3sqrt(1-x)(-2-x)+C`

`=-2/3sqrt(1-x)(2+x)+C`

Integrate by parts:

`int (xdx)/sqrt(1-x) = int x d(-2sqrt(1-x))`

`int (xdx)/sqrt(1-x) = -2x sqrt(1-x) + 2 int sqrt(1-x)dx`

`int (xdx)/sqrt(1-x) = -2x sqrt(1-x) - 2 int (1-x)^(1/2)d(1-x)`

`int (xdx)/sqrt(1-x) = -2x sqrt(1-x) - 4/3 (1-x)^(3/2) + C`

`int (xdx)/sqrt(1-x) = -2x sqrt(1-x) - 4/3 (1-x)sqrt(1-x) + C`

`int (xdx)/sqrt(1-x) = -sqrt(1-x)(2x+ 4/3 (1-x)) + C`

`int (xdx)/sqrt(1-x) = -sqrt(1-x)(2/3x+ 4/3 ) + C`

`int (xdx)/sqrt(1-x) = -(2(x+2)sqrt(1-x))/3 + C`

Let, `I=intx/sqrt(1-x)dx=-int(-x)/sqrt(1-x)dx`,

`=-int{(1-x)-1}/sqrt(1-x)dx`,

`=-int{(1-x)/sqrt(1-x)-1/sqrt(1-x)}dx`,

`=-int{sqrt(1-x)-1/sqrt(1-x)}dx`,

`=-int(1-x)^(1/2)dx+int(1-x)^(-1/2)dx`.

Recall that,

`intf(x)dx=F(x)+C rArr intf(ax+b)dx=1/aF(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`.