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Question:

$\frac{1}{\sqrt{x+a}+\sqrt{(x+b)}}$

Solution:

$\frac{1}{\sqrt{x+a}+\sqrt{x+b}}=\frac{1}{\sqrt{x+a}+\sqrt{x+b}} \times \frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}}$

$=\frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)}$

$=\frac{(\sqrt{x+a}-\sqrt{x+b})}{a-b}$

$\Rightarrow \int \frac{1}{\sqrt{x+a}-\sqrt{x+b}} d x=\frac{1}{a-b} \int(\sqrt{x+a}-\sqrt{x+b}) d x$

$=\frac{1}{(a-b)}\left[\frac{(x+a)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(x+b)^{\frac{3}{2}}}{\frac{3}{2}}\right]$

$=\frac{2}{3(a-b)}\left[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{3}{2}}\right]+\mathrm{C}$

 

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