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 $\frac{\large 1}{\large x(log x)^m},x\;>\;0$

$\begin{array}{1 1} \frac{(\log x)^{1-m}}{1-m}+c \\\frac{(\log x)^{1+m}}{1-m}+c \\ \frac{(\log x)^{1-m}}{1+m}+c \\ \frac{(\log x)^{1+m}}{1+m}+c\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Method of substitution :
  • Given f(x)dx can be transformed into another form by changing independent variable x to t by substituting x=g(t).
  • Consider $I=\int f(x)dx.$
  • Put x=g(t) so that $\frac{dx}{dt}=g'(t).$
  • dx=g'(t)dt.
  • Thus $I=\int f(g(t).g'(t))dt.$
Given $I=\int\frac{1}{x(log x)^m}dx,x>0$.
 
Let $log x=t$.
 
$\;\;\;\frac{1}{x}dx=dt$.
 
Now substituting for x and dx we get,
 
$I=\int\frac{dt}{t^m}$.
 
 
On integrating we get,
$\;\;\;=\frac{t^{-m+1}}{-m+1}+c.$
 
 
Substituting for m we get,
 
$\;\;\;=\frac{(log x)^{1-m}}{1-m}+c.$
 
Hence $\int\frac{1}{x(log x)^m}dx=\frac{(log x)^{1-m}}{1-m}+c.$
 
 
 

 

answered Jan 28, 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
...