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Integrate the functions\[\frac{1}{\sqrt{x^2+2x+2}}\]

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

1 Answer

Toolbox:
  • $\int\frac{dx}{\sqrt{x^2+a^2}}=\int log\mid x+\sqrt{x^2+a^2}\mid+c.$
Given:$I=\int\frac{dx}{\sqrt{x^2+2x+2}}dx.$
 
$\sqrt{x^2+2x+2}$ can be written as $\sqrt{(x+1)^2+1^2}.$
 
$I=\int\frac{dx}{\sqrt {(x+1)^2+1^2}}$.
 
On integrating we get,
 
$\;\;\;=log\mid(x+1)+\sqrt{x^2+2x+2}\mid+c.$

 

answered Feb 4, 2013 by sreemathi.v
 
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