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

Choose the correct answer in $\Large \int \normalsize \sqrt{1+x^2}dx$ is equal to \[\begin{array}{l}(A)\;\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}log\LARGE\mid\normalsize\bigg(x+\sqrt{1+x^2}\bigg)\LARGE\mid\normalsize+C \\(B)\;\frac{2}{3}(1+x^2)^\frac{3}{2}+C\qquad \\ (C)\;\frac{2}{3}x(1+x^2)^\frac{3}{2}+ C \\ (D)\;\frac{x^2}{2}\sqrt{1+x^2}+\frac{1}{2}x^2log\LARGE\mid\normalsize x+\sqrt{1+x^2}\LARGE\mid\normalsize+C \end{array}\]

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\int \sqrt{a^2+x^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}log |x+\sqrt{x^2+a^2}$
       Given $ I=\int \sqrt{1+x^2} dx$
 
This is of the form $\int \sqrt{a^2+x^2}dx$
 
                     $=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}log |x+\sqrt{x^2+a^2}|+c$
 
            Here x=x and a=1
 
Therefore $\int \sqrt{1+x^2}dx=\frac{x}{2}\sqrt {x^2+1}+\frac{1}{2}log |x+\sqrt{x^2+1}|$
 
Hence A is the correct answer

 

answered Feb 8, 2013 by meena.p
 

Related questions

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
...