logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Integrals
0 votes

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} $

Can you answer this question?
 
 

1 Answer

0 votes
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
 
Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...