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

Prove the following\[\int\limits_1^3\frac{dx}{x^2(x+1)}=\frac{2}{3}+log\frac{2}{3}\]

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • If the given integral function in a rational function of the form $\large\frac{A}{(x+1)(x+2)^2}=\frac{A}{(x+_1)}+\frac{B}{(x+2)}+\frac{c}{(x+2)^2}$
  • (ii)$\int \frac{dx}{(x+a)} =\log (x+a) +c$
  • (iii)$ \int \limits_a^b f(x)dx=F(b)-F(a)$
$\int \limits_1^3 \frac{dx}{x^2(x+1)}=\frac{2}{3}+log \frac{2}{3}$
 
given $I=\int \limits_1^3 \frac{dx}{x^2(x+1)}$
 
This is a proper rational function,hence it can be resolved into its partial function as
 
$\large\frac{1}{x^2(x+1)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}$
 
$1=Ax(x+1)+B(x+1)+B+Cx^2$
 
$=Ax^2+A+Bx+B+Cx^2$
 
Equating the coefficient of $x^2,$
 
$0=A+C$ ----------(1)
 
Equating the coefficient of $x$
 
$0=A+B$----------(2)
 
Equating the constant term,
 
$1=B \qquad Hence\; B=1$
 
Substituting for B in equ (2) we get,
 
$ A+1=0 => A =-1$
 
Substituting for A in equ (1)
 
$A+C=0$
 
$-1+C=0 => C=1$
 
Hence $A =-1,B=1,C=1$
 
Hence on substituting for A,B and C we get
 
$\frac{1}{x^2(x+1)}=-\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x+1}$
 
Therefore $I=\int \limits_1^3 \frac{-1}{x} dx+\int \limits_1^3 \frac{1}{x^2}dx+\int \limits_1^3 \frac{a}{x+1} dx$
 
On integrating we get
 
$I=\bigg[-log x\bigg]_1^3+\bigg[\frac{x^{-2+1}}{-2+1}\bigg]_1^3+\bigg[log (x+1)\bigg]_1^3dx$
 
$=[-log x]_1^3-[\frac{1}{x}]_1^3+[log (x+1)]_1^3$
 
On applying limits,
 
$-(log 3- log 1)-(\frac{1}{3}-1)+[log[(3+1)-log 2]]$
 
But $log (a/b) =log a-log b$
 
Therefore $-(log 3/1)+\frac{2}{3}+log \frac{4}{2}$
 
$=\frac{2}{3}+\log (\frac{4}{2} /\frac{3}{1})$
 
$=\frac{2}{3}+log (\frac{2}{3})$
 
Hence $I=\frac{2}{3}+log (\frac{2}{3})$
 
Hence proved

 

 

answered Mar 7, 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
...