Question
Evaluate the following integrals:
$\int\frac{1}{1+\sqrt{\text{x}}}\text{dx}$

Answer

Let $\text{I}=\int\frac{1}{1+\sqrt{\text{x}}}\text{dx}\ \ ....(1)$
Let $\text{x}=\text{t}^2$ then,
$\text{dx}=\text{d}(\text{t}^2)$
$\Rightarrow\text{dx}=2\text{t}\text{ dt}$
Putting $\text{x}=\text{t}^2$ and $\text{dx}=2\text{t}\text{ dt}$ in equation (1), we get
$\text{I}=\int\frac{2\text{t}}{1+\sqrt{\text{t}^2}}\text{dt}$
$=\int\frac{2\text{t}}{1+\text{t}}\text{dt} $
$=2\int\frac{\text{t}}{1+\text{t}}\text{dt}$
$=2\int\frac{1+\text{t}-1}{1+\text{t}}\text{dt}$
$=2\int\Big[\frac{1+\text{t}}{1+\text{t}}-\frac{1}{1+\text{t}}\Big]\text{dt}$
$=2\int\text{dt}-2\int\frac{1}{1+\text{t}}\text{dt} $
$=2\text{t}-2\log(1+\text{t})+\text{C}$
$=2\sqrt{\text{x}}-2\log(1+\sqrt{\text{x}})+\text{C}$
$\therefore\text{ I}=2\sqrt{\text{x}}-2\log(1+\sqrt{\text{x}})+\text{C}$

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