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,

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

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

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

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