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

Question

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

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Answer 1 · 2018-04-14T12:21:05

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

Here,

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

Let, #x+1=u=>x=u-1=>dx=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#