How do you integrate #int 1/(sqrtx(sqrtx+1)^2)# using substitution?

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Answer 1 · 2016-12-30T12:33:59

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

We need

#intx^ndx=x^(n+1)/(n+1)+C (n!=-1)#

We use the substitution

#u=sqrtx+1#

#du=dx/(2sqrtx)#

Therefore,

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

#=2*-1/u#

#=-2/(sqrtx+1)+C#