How do you integrate #x/(x+1)^3#?

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Answer 1 · 2018-07-13T09:30:19

The answer is #=-1/(x+1)+1/(2(x+1)^2)+C#

Perform this integral by substitution

Let #u=x+1#, #=>0#, #du=dx#

The integral is

#I=int(xdx)/(x+1)^3=int((u-1)du)/u^3#

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

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

#=-1/(x+1)+1/(2(x+1)^2)+C#

Answer 2 · 2018-07-13T09:46:24

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

Here,

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

Subst . #color(blue)(x+1=u=>x=u-1=>dx=du#

So,

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

#=int[1/u^2-1/u^3]du#

#=int[u^-2 -u^-3 ]du#

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

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

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

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

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

Subst. back #color(blue)(u=x+1#

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

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