How do you evaluate the integral $\int \frac{1}{1 + \sqrt{x}}$ $int 1/(1+sqrtx)$ ?

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Answer 1 · 2016-12-23T11:23:14

The answer is $= 2 \sqrt{x} - 2 ln (\sqrt{x} + 1) + C$ $=2sqrtx-2ln(sqrtx+1)+ C$

We can perform this integral by substitution

Let $u = \sqrt{x}$ $u=sqrtx$ , $\Rightarrow$ $=>$ , $d u = \frac{d x}{2 \sqrt{x}}$ $du=(dx)/(2sqrtx)$

So,

$\int \frac{d x}{\sqrt{x} + 1} = \int \frac{2 \sqrt{x} d u}{u + 1}$ $intdx/(sqrtx+1)=int(2sqrtxdu)/(u+1)$

$= 2 \int \frac{u d u}{u + 1}$ $=2int(udu)/(u+1)$

$\frac{u}{u + 1} = 1 - \frac{1}{u + 1}$ $u/(u+1)=1-1/(u+1)$

Therefore,

$\int \frac{d x}{\sqrt{x} + 1} = 2 \int (1 - \frac{1}{u + 1}) d u$ $intdx/(sqrtx+1)=2int(1-1/(u+1))du$

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

$= 2 \sqrt{x} - 2 ln (\sqrt{x} + 1) + C$ $=2sqrtx-2ln(sqrtx+1)+ C$

How do you evaluate the integral int 1/(1+sqrtx)∫11+√xint 1/(1+sqrtx)?