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 rational functions\[\frac{3x-1}{(x-1)(x-2)(x-3)}\]

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $(i)\;$Form of the rational function\[\frac{px+q}{(x-a)(x-b)(x-c)},a\neq b\neq c\]
  • $(ii)\;$Form of the partial function\[\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{(x-c)}\]
Given $I=\int\frac{3x-1}{(x-1)(x-2)(x-3)}.$
 
This can be written as:
 
$\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x-2)}+\frac{C}{(x-3)}.$
 
$\Rightarrow 3x-1=A(x-2)(x-3)+B(x-1)(x-3)+c(x-1)(x-2)$.
 
Equating the coefficients of $x^2$,
 
0=A+B+C-----(1)
 
equating the coefficients of x,
 
3=-5A-5B-3C----(2)
 
Equating the constant terms,
 
-1=6A+3B+2C------(3)
 
Let us solve for A,B and C.
 
Multiply equ(1) by 5 and add with equ(2)
5A+5B+5C=0
-5A-4B-3C=3
____________________
B+2C=3------(4)
 
Multiply equ(1) by 6 and subtract from equ(3)
6A+3B+2C=-1
6A+6B+6C=0
______________________
-3B-4C=-1-------(5)
 
Multiply equ(4) by 3 and add equ(5)
3B+6C=9
-3B-4C=-1
________________
2C=8$\Rightarrow C=4.$
Substituting for C in equ(4) we get,
 
B+8=3$\Rightarrow B=-5.$
 
Substituting for B and C in equ(1)
 
A-5+4=0.
 
$\rightarrow A=1.$
 
Hence A=1,B=-5 and C=4.
 
Now substituting for A,B and C in I we get,
 
$I=\int\frac{1}{(x+1)}+\frac{(-5)}{(x-2)}+\frac{4}{(x-3)}dx.$
 
On separating the terms we get,
 
$I=\int\frac{1}{x-1}dx-\int\frac{5}{(x-2)}dx+\int\frac{4}{(x-3)}dx.$
 
On integrating we get,
 
$\;\;\;=log|x-1|-5log|x-2|+4log|x-3|+c.$
 
Hence $ I=log|x-1|-5log|x-2|+4log|x-3|+c.$

 

answered Feb 5, 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
...