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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the rational functions\[\frac{2x-3}{(x^2-1)(2x+3)}\]

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Toolbox:
  • $(i)\;$Form of the rational function\[\frac{px+q}{(x-a)(x-b)(x-c)}\]
  • $\;$Form of the partial function\[\frac{A}{(x-a)}+\frac{B}{(x-b}+\frac{C}{(x-c)}\]
  • $(ii)\;\int\frac{dx}{(x-a)}=log|x-a|+c.$
Given $I=\int\frac{2x-3}{(x^2-1)^(2x+3)}dx=\int\frac{2x-3}{(x-1)(x+1)(2x+3)}.$
 
$\frac{2x-3}{(x-1)(x+2)(2x+3)}=\frac{A}{(x-1)}+\frac{B}{(x+1)}+\frac{C}{(2x+3)}.$
 
2x-3=A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1).$
 
$2x-3=A(2x^2+5x+3)+B(2x^2+x-3)+C(x^2-1).$
 
Equating the coefficients of $x^2$,
 
0=2A+2B+C-----(1)
 
Equating the coefficients of x,
 
2=5A+B------(2)
 
Equating the constant terms,
 
-3=3A-3B-C-------(3)
 
Multiply equ(2) by 2 and subtract from equ(1)
 
Subtract equ(1) and equ(2)
 
2A+2B+C=0
10A+2B=4
_________________
8A+C=-4-------(4)
Multiply equ(2) by 3 and add it with equ(3),
3A-3B-C=-3
15A+3B=6
___________________
18A-C=3-----(5)
Now add equ(4) and equ(5)
18A-C=3
-8A+C=-4
________________
10A=-1
$A=\frac{-1}{10}$
 
Substituting for A in equ(5) we get,
 
18(-1/10)-C=3
 
$\frac{-9}{5}-C=3$
 
$\Rightarrow -C=3+\frac{9}{5}.$
 
C=$\frac{-24}{5}.$
 
substituting for A in equ(2)
 
5A+B=2.
 
5(-1/10)+B=2
 
B=2+$\frac{1}{2}$
 
$B=\frac{5}{2}$
 
Hence $A=-1/10,B=5/2 and c=-24/5.
 
Now substituting the values of A,B and C we get,
 
$\frac{2x-3}{(x-1)(x+1)(2x+3)}=\frac{-1}{10(x-1)}+\frac{5}{2(x-1)}-\frac{24}{5(2x+3)}$
 
Therefore $I=\frac{-1}{10}\int{dx}{(x-1)}+\frac{5}{2}\int\frac{dx}{(x+1)}-\frac{24}{5}\int\frac{dx}{2x+3}.$
 
Put 2x+3=t.
 
Therefore 2xdx=dt$\Rightarrow dx=\frac{dt}{2}.$
 
Substituting this
 
$I=\frac{-1}{10}\int\frac{dx}{(x-1)}+\frac{5}{2}\int\frac{dx}{(x+1)}-\frac{24}{5}\times\frac{1}{2}\int\frac{dt}{t}.$
 
On integrating we get,
 
$\;\;\;=\frac{-1}{10}log|x-1|+\frac{5}{2}log|x+1|-\frac{12}{5}log|t|.$
 
Substituting for t,
 
$\int\frac{2x-3}{(x^2-1)(2x+3)}dx=\frac{-1}{10}log|x-1|+\frac{5}{2}log|x+1|-\frac{12}{5}log|2x+3|+c.$
answered Feb 6, 2013 by sreemathi.v
edited Apr 9, 2014 by balaji.thirumalai
 
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