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Integrate the function\[\frac{1}{\sqrt{x+a}+\sqrt{x+b}}\]

1 Answer

Toolbox:
  • $=\int \sqrt{x+a}=\frac{(x+a)^{3/2}}{3/2}=\frac{2}{3}(x+a)^{3/2}$
Given:$\frac{1}{\sqrt{x+a}+\sqrt{x+b}}$
 
Multiply and divide by its conjujate $\sqrt {x+a}-\sqrt {x+b}$
 
$=\large \int \frac{\sqrt {x+a}-\sqrt {x+b}}{(\sqrt {x+a}+\sqrt {x+b})(\sqrt {x+a}-\sqrt {x+b})}dx$
 
$=\large \int \frac{\sqrt {x+a}-\sqrt {x+b}}{(\sqrt {x+a})^2-(\sqrt {x+b})^2}dx$
 
$=\large \int \frac{\sqrt {x+a}-\sqrt {x+b}}{(x+a-x-b)}dx$
 
$=\large \int \frac{\sqrt {x+a}-\sqrt {x+b}}{a-b}dx$
 
On seperating the terms:
 
$=\frac{1}{a-b}\bigg[\int \sqrt{x+a}\;dx-\int \sqrt {x+b}\;dx\bigg]$
 
On integreating we get,
 
$\frac{1}{(a-b)}\bigg\{\bigg[\frac{(x+a)^{3/2}}{3/2}\bigg]-\bigg[\frac{(x+b)^{3/2}}{3/2}\bigg]\bigg\}$
 
$=\frac{1}{(a-b)}\frac{2}{3}\bigg[(x+a)^{3/2}-(x+b)^{3/2}\bigg]$
 
$I=\frac{2}{3(a-b)}\bigg[(x+a)^{3/2}-(x+b)^{3/2}\bigg]$

 

 

answered Feb 15, 2013 by meena.p
 
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