Integral of int(x-1)/(x+1)^3x1(x+1)3 dx=?

int(x-1)/(x+1)^3 dx x1(x+1)3dx

1 Answer
Apr 14, 2018

I=-x/(x+1)^2+cI=x(x+1)2+c

Explanation:

Here,

I=int(x-1)/(x+1)^3dxI=x1(x+1)3dx

Let, x+1=u=>x=u-1=>dx=dux+1=ux=u1dx=du

:.I=int(u-1-1)/u^3du

=int(u/u^3-2/u^3)du

=int(u^-2-2u^-3)du

=(u^(-2+1))/(-2+1)-2((u^(-3+1))/(-3+1))+c

=-u^-1-2(u^-2/-2)+c

=-1/u+u^-2+c

I=1/u^2-1/u+c=(1-u)/u^2+c

=(1-(x+1))/(x+1)^2...to[as,u=x+1 ]

I=-x/(x+1)^2+c