What is the integration of (xdx)/sqrt(1-x)xdx1x??

3 Answers

-2/3sqrt(1-x)(2+x)+C231x(2+x)+C

Explanation:

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

or, u^2=1-xu2=1x

or, x=1-u^2x=1u2

or, dx=-2ududx=2udu

Now, int (xdx)/(sqrt(1-x))=int (1-u^2)(-2udu)/u=int 2u^2du -int 2duxdx1x=(1u2)2uduu=2u2du2du

Now, int 2u^2 du -int 2du2u2du2du

=(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=2u332(u)+C=23u(u23)+C=231x{(1x)3}+C=231x(2x)+C

=-2/3sqrt(1-x)(2+x)+C=231x(2+x)+C

Mar 6, 2018

int (xdx)/sqrt(1-x) = -(2(x+2)sqrt(1-x))/3 + Cxdx1x=2(x+2)1x3+C

Explanation:

Integrate by parts:

int (xdx)/sqrt(1-x) = int x d(-2sqrt(1-x))xdx1x=xd(21x)

int (xdx)/sqrt(1-x) = -2x sqrt(1-x) + 2 int sqrt(1-x)dxxdx1x=2x1x+21xdx

int (xdx)/sqrt(1-x) = -2x sqrt(1-x) - 2 int (1-x)^(1/2)d(1-x)xdx1x=2x1x2(1x)12d(1x)

int (xdx)/sqrt(1-x) = -2x sqrt(1-x) - 4/3 (1-x)^(3/2) + Cxdx1x=2x1x43(1x)32+C

int (xdx)/sqrt(1-x) = -2x sqrt(1-x) - 4/3 (1-x)sqrt(1-x) + Cxdx1x=2x1x43(1x)1x+C

int (xdx)/sqrt(1-x) = -sqrt(1-x)(2x+ 4/3 (1-x)) + Cxdx1x=1x(2x+43(1x))+C

int (xdx)/sqrt(1-x) = -sqrt(1-x)(2/3x+ 4/3 ) + Cxdx1x=1x(23x+43)+C

int (xdx)/sqrt(1-x) = -(2(x+2)sqrt(1-x))/3 + Cxdx1x=2(x+2)1x3+C

Mar 6, 2018

-2/3(2+x)sqrt(1-x)+C23(2+x)1x+C.

Explanation:

Let, I=intx/sqrt(1-x)dx=-int(-x)/sqrt(1-x)dxI=x1xdx=x1xdx,

=-int{(1-x)-1}/sqrt(1-x)dx=(1x)11xdx,

=-int{(1-x)/sqrt(1-x)-1/sqrt(1-x)}dx={1x1x11x}dx,

=-int{sqrt(1-x)-1/sqrt(1-x)}dx={1x11x}dx,

=-int(1-x)^(1/2)dx+int(1-x)^(-1/2)dx=(1x)12dx+(1x)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)+Cf(ax+b)dx=1aF(ax+b)+K,(a0)

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.