How do you integrate #int -x/((x+1)-sqrt(x+1))dx#?

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Answer 1 · 2017-02-20T15:34:10

#-x-2sqrt(x+1)+C#

In order to integrate this, we need to use a method called u-substitution, but a good idea is to remove all constants out of the integral first.

#int -x/((x+1)-sqrt(x+1))dx=-int x/(x-sqrt(x+1)+1)dx#

Let #u=sqrt(x+1)#

#(du)/dx=1/(2sqrt(x+1))#

#dx=2sqrt(x+1) xxdu=2u# #du#

#x=u^2-1#

The integral can now be rewritten in terms of #u#

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

#-2intu+1# #du=-2(1/2u^2+u)+C=-u^2-2u+C=#

#-(x+1)-2(sqrt(x+1))+C=-x-1-2sqrt(x+1)+C#

Since #1# is a constant, we can add it to #C#, leaving the answer as:

#-x-2sqrt(x+1)+C#