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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[\sqrt{x^2+4x+6}\]

$\begin{array}{1 1} \large \frac{(x+2)(\sqrt{x^2+4x+6})}{2}+\log \mid (x+2)+\sqrt{x^2+4x+6} \mid +c \\\large \frac{(x+2)(\sqrt{x^2+4x+6})}{2}+\log \mid (x+2)-\sqrt{x^2+4x+6} \mid +c \\\large \frac{(x+2)(\sqrt{2x+4})}{2}+\log \mid (x+2)+\sqrt{x^2+4x+6} \mid +c \\\large \frac{(x+2)(\sqrt{2x+4})}{2}+\log \mid (x+2)+\sqrt{2x+4} \mid +c\end{array} $

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  • $\int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac {a^2}{2}log|x+\sqrt{x^2+a^2}|+c$
Given $I=\sqrt{x^2+4x+6}dx$
 
we can split $I=\sqrt{x^2+4x+6}\;as\;\sqrt{(x^2+4x+4)-4+6}$
 
$\sqrt{(x+2)^2-4+6}=\sqrt{(x+2)^2+2}$
 
Hence $I=\sqrt{(x+2)^2+(\sqrt 2})^2$
 
This is of the form $\sqrt{x^2+a^2}=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}log|(x+2)+\sqrt{x^2+a^2}|+c$
 
Here x=(x+2) and $a=\sqrt 2.$
 
Therefore $I=\frac{(x+2)}{2}\sqrt{x^2+4x+6}+\frac{(\sqrt 2)^2}{2} log|x+\sqrt{x^2+4x+6}|+c$
 
$I=\frac{(x+2)}{2}\sqrt{x^2+4x+6}+log|(x+2)+\sqrt{x^2+4x+6}|+c$

 

answered Feb 7, 2013 by meena.p
 
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