Integrate the function in Exercise:1 x+a + x+b - Vidyadip

Question

Integrate the function in Exercise:$\frac{1}{\sqrt{\text{x}+\text{a}}+\sqrt{\text{x}+\text{b}}}$

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Answer

$\frac{1}{\sqrt{\text{x}+\text{a}}+\sqrt{\text{x}+\text{b}}}=\frac{1}{\sqrt{\text{x}+\text{a}}+\sqrt{\text{x}+\text{b}}}\times\frac{\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}}{\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}}$
$=\frac{\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}}{(\text{x}+\text{a)}-(\text{x}+\text{b)}}$
$=\frac{\big(\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}\big)}{\text{(a}-\text{b)}}$
$\Rightarrow\int\frac{1}{\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}}\text{dx}=\frac{1}{\text{a}-\text{b}}\int\Big(\sqrt{\text{x}+\text{a}}-\sqrt{\text{x}+\text{b}}\Big)\text{dx}$
$=\frac{1}{\text{(a}-\text{b)}}\left[\frac{\text{(x}+\text{a)}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{\text{(x}+\text{b)}^{\frac{3}{2}}}{\frac{3}{2}}\right]$
$=\frac{2}{3\text{(a}-\text{b)}}\left[\text{(x}+\text{a)}^{\frac{3}{2}}-\text{(x}+\text{b)}^{\frac{3}{2}}\right]+\text{C}$

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