Evaluate the following integrals:


Evaluate the following integrals:

$\int \frac{1}{1+\sqrt{x}} d x$



$\mathrm{d}(\mathrm{x})=2 \mathrm{t} \cdot \mathrm{dt}$

$\mathrm{dx}=2 \mathrm{t} \cdot \mathrm{dt}$

Substituting $\mathrm{t}$ and dt we get

$\Rightarrow \int \frac{2 t \cdot d t}{1+t}$

$\Rightarrow 2 \int \frac{t d t}{1+t}$

Add and subtract 1 from numerator

$\Rightarrow 2 \int \frac{t+1-1}{1+t} d t$

$\Rightarrow 2\left(\int \frac{t+1}{t+1} d t-\int \frac{1}{1+t} d t\right)$

$\Rightarrow 2\left(\int d t-\int \frac{1}{1+t} d t\right)$

$\Rightarrow 2(\mathrm{t}-\ln |1+\mathrm{t}|)$

But $t=\sqrt{x}$

$\Rightarrow 2(\sqrt{x}-\ln |1+\sqrt{x}|)+c$

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