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 function\[\sqrt{1+\frac{x^2}{9}}\]

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac {a^2}{2}log |x+\sqrt{x^2+a^2}|+c$
Given $I=\int \sqrt{1+\frac{x^2}{9}}dx$
 
$ 1+\frac{x^2}{9}=\frac{9+x^2}{9}$
 
$=\frac{1}{9}(9+x^2)$
 
Therefore $I=\sqrt{\frac{1}{9}(x^2+9)}dx$
 
$=\frac{1}{3}\sqrt{x^2+9}dx$
 
Clearly this is of the form $\sqrt{x^2+a^2}=$
 
$=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}log|x+\sqrt{x^2+a^2}|+c$
 
Here in the place of x we have x and in the place of a we have 3
 
Therefore $\frac{1}{3}\int \sqrt{x^2+9} dx=\frac{1}{3}[\frac{x}{2}\sqrt{x^2+9}+\frac{9}{2} log |(x+\sqrt{x^2+9}|+c$
 
$=\frac{x}{6}\sqrt{x^2+9}+\frac{3}{2}log|x+\sqrt{x^2+9}|+c$

 

answered Feb 8, 2013 by meena.p
 
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
...