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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[\sqrt{x^2+4x-5}\]

$\begin{array}{1 1} \frac{(x+2) (\sqrt{x^2+4x-5})}{2}-\frac{9}{2} \log \mid (x+2)+\sqrt{x^2+4x-5}\mid+c \\ \frac{(x+2) (\sqrt{x^2+4x-5})}{2}+\frac{9}{2} \log \mid (x+2)+\sqrt{x^2+4x-5}\mid+c \\ \frac{(x+2) (\sqrt{x^2+4x-5})}{2}-\frac{9}{2} \log \mid \sqrt{x^2+4x-5}\mid+c \\ \frac{(x+2) (\sqrt{x^2+4x-5})}{2}+\frac{9}{2} \log \mid (x+\sqrt{x^2+4x-5}\mid+c\end{array} $

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1 Answer

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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 $I=\int \sqrt{x^2+4x-5}dx$
 
we can split $(x^2+4x-5)\;as\;x^2+4x+4-4-5$
 
                     $=(x+2)^2-9$
                     
                     $=(x+2)^2-(3)^2$
 
                     $I=\int \sqrt{(x+2)^2-(3)^2 }dx$
 
This is of the form $\sqrt{x^2-a^2}$
 
                     $=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}log|x+\sqrt{x^2-a^2}|+c$
 
         where x= x+2 and a=3.
 
Therefore $\int \sqrt{(x^2)^2-(3)^2}=\frac{(x+2)}{2}\sqrt{x^2+4x-5}-\frac{9}{2} |(x+2)+\sqrt{x^2+4x-5}|+c$
 
                      $=\frac{(x+2)}{2}\sqrt{x^2+4x-5}-\frac{9}{2}|(x+2)+\sqrt{x^2+4x+1}|+c$
answered Feb 7, 2013 by meena.p
edited Apr 11 by meena.p
 
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