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 answers in $\int\frac{dx}{e^x+e^{-x}}$ is equal to

$\begin{array}{1 1} (A)\;\tan^{-1}(e^x)+C \\ (B)\;\tan^{-1}(e^{-x})+C\\C\;log(e^x-e^{-x})+C \\ (D)\;log(e^x+e^{-x})+C \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • (i)In a function $ \int f(x)dx$ if f(x)=t, then $f'(x)dx=dt$ Therefore $\int f(x)dx=\int t.dt$
  • (ii) $\int \frac{dx}{x^2+a^3}=\frac{1}{a} \tan ^{-1}(x/a)+c$
Given $I=\frac{dx}{e^x+e^{-x}}$
 
multiply and divide by $e^x$
 
$I=\int \frac{e^x}{e^x(e^x+e^{-x})}dx$
 
$I=\int\frac{e^x}{e^x(e^{2x}+1)}dx$
 
Let $e^2x=t$ on differentiating w.r.t x,
 
$e^xdx=dt$
 
Therefore $I=\int \frac{dt}{1+t^2}$
 
This is of the form $\int \frac{dx}{x^2+a^2}=\frac{1}{a} \tan ^{-1}(x/a)+c$
 
Therefore on integrating we get
 
$I=\tan ^{-1}(t)+c$
 
substituting for t we get,
 
$I=\tan ^{-1}(e^x)+c$
 
Hence the correct answer is A

 

 

answered Mar 13, 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
...