Question
Evaluate : $\int_1^3 \frac{1}{\sqrt{2+x}+\sqrt{x}} \cdot d x$

Answer

$\begin{aligned} & =\int_1^3\left(\frac{1}{\sqrt{2+x}+\sqrt{x}}\right)\left(\frac{\sqrt{2+x}-\sqrt{x}}{\sqrt{2+x}-\sqrt{x}}\right) \cdot d x \\ & =\int_1^3\left(\frac{\sqrt{2+x}-\sqrt{x}}{2+x-x}\right) \cdot d x \\ & =\frac{1}{2} \cdot \int_1^3(\sqrt{2+x}-\sqrt{x}) \cdot d x \\ & =\frac{1}{2} \cdot\left[\frac{(2+x)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_1^3\end{aligned}$
$\begin{aligned} & =\frac{1}{3} \cdot\left[(2+x)^{\frac{3}{2}}-(x)^{\frac{3}{2}}\right]_1^3 \\ & =\frac{1}{3}\left\{\left[(2+3)^{\frac{3}{2}}-(3)^{\frac{3}{2}}\right]-\left[(2+1)^{\frac{3}{2}}-(1)^{\frac{3}{2}}\right]\right\} \\ & =\frac{1}{3}\left\{5^{\frac{3}{2}}-3^{\frac{3}{2}}-3^{\frac{3}{2}}+1^{\frac{3}{2}}\right\} \\ & =\frac{1}{3}\left\{5^{\frac{3}{2}}-2(3)^{\frac{3}{2}}+1\right\} \\ & \therefore \int_1^3 \frac{1}{\sqrt{2+x}+\sqrt{x}} d x=\frac{1}{3}\left[5^{\frac{3}{2}}-2(3)^{\frac{3}{2}}+1\right]\end{aligned}$

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