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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $\sqrt{x^2+4x+1}$

$\begin{array}{1 1} \large \frac{(x+2)(\sqrt{x^2+4x+1}}{2}-\frac{3}{2} \log|x+2+\sqrt{x^2+4x+1}|+c \\\large \frac{(x+2)(\sqrt{x^2+4x+1}}{2}+\frac{3}{2} \log|x+2+\sqrt{x^2+4x+1}|+c \\ \large \frac{(x+2)(\sqrt{x^2+4x+1}}{2}-\frac{3}{2} \log|x+2-\sqrt{x^2+4x+1}|+c \\\large \frac{(x-2)(\sqrt{x^2+4x+1}}{2}-\frac{1}{2} \log|x+2+\sqrt{x^2+4x+1}|+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 $I=\sqrt{x^2+4x+1}dx$
 
we can split $I=\sqrt{x^2+4x+1}\;as\;\sqrt{(x^2+4x+4)-4+1}$
 
$I=\sqrt{(x+2)^2+(\sqrt 3})^2$
 
This is of the form $\sqrt{x^2-a^2}$
 
$\sqrt{x^2-a^2}=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}log|x+\sqrt{x^2-a^2}|+c$
 
Here x=(x+2) and $a=\sqrt 3.$
 
Now substituting for x and a we get
 
$\int \sqrt{x^2+4x+1}dx=\frac{(x+2)}{2}\sqrt{x^2+4x+1}-\frac{3}{2} log|x+\sqrt{x^2+4x+1}|+c$
 
$=\frac{(x+2)}{2}\sqrt{x^2+4x+1}-\frac{3}{2}log|x+\sqrt{x^2+4x+1}|+c$

 

answered Feb 7, 2013 by meena.p
 
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