How do you evaluate the integral #int 1/(1+sqrtx)#?

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Answer 1 · 2016-12-23T11:23:14

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

We can perform this integral by substitution

Let #u=sqrtx#, #=>#, #du=(dx)/(2sqrtx)#

So,

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

#=2int(udu)/(u+1)#

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

Therefore,

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

#=2u-2ln(u+1)#

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