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

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Toolbox:
  • $(i)\;$Form of the rational function\[\frac{px+q}{(x^2+a)(x^2+b)}\]
  • $\;$Form of the partial function\[\frac{Ax+B}{(x^2+a)}+\frac{Cx+D}{(x^2+b)}\]
  • $(ii)\int\frac{dx}{x}=log|x|+c.$
Given$I=\int \frac{2x}{(x^2+1)(x^2+3)}$
 
Resolving into partial functions
 
Let $\frac{2x}{(x^2+1)(x^2+3)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+3}.$
 
$\Rightarrow 2x=(Ax+B)(x^2+3)+(Cx+D)(x^2+1).$
 
Equating the coefficients of $x^3$.
 
0=A+C----(1)
 
Equating the coefficients of $x^2$,
 
0=B+D----(2)
 
Equating the coefficients of $x$,
 
2=3A+C----(3)
 
Equating the constant terms
 
0=3B+D----(4)
 
Consider equ(1) and equ(3)
 
On subtracting them we get,
 
A+C=0
3A+C=2
________________
-2A=-2$\Rightarrow A=1.$
Substitute for A in eq(1)
 
A+C=0.
 
1+C=0$\Rightarrow C=-1.$
 
Consider eq(2) and eq(4),
 
On subtracting them we get,
 
B+D=0.
 
3B+D=0.
 
since both the equations are equated to 0,B and D are 0.
 
Hence A=1 and C=-1.
 
Now substituting for A and C in I we get,
 
$\frac{2x}{(x^2+1)(x^2+3)}=\frac{x}{x^2+1}-\frac{x}{x^2+3}.$
 
For integrating we can use the substitution method,
 
$I=\int\frac{x}{x^2+1}dx-\int\frac{x}{x^2+3}dx.$
 
Let $x^2=t.$
 
On differentiating we get,
 
2xdx=dt$\Rightarrow xdx=\frac{dt}{2}.$
 
Similarly let $x^2+3=u$,on differentiating we get,
 
2xdx=du$\Rightarrow xdx=\frac{du}{2}.$
 
Now substituting these values we get,
 
$\;\;\;=\int\frac{dt/2}{t}-\int\frac{du/2}{u}.$
 
$\;\;\;=\frac{1}{2}\int\frac{dt}{t}-\frac{1}{2}\int\frac{du}{u}.$
 
On integrating we get,
 
$\;\;\;=\frac{1}{2}log|t|-\frac{1}{2}log|u|.$
 
$log a-log b=log|\frac{a}{b}|$,similarly
 
$\;\;\;=\frac{1}{2}log\frac{|t|}{|u|}+c.$
 
Substituting back for t and u we get,
 
$\int\frac{2x}{(x^2+1)(x^2+3)}=\frac{1}{2}log\frac{(x^2+1)}{(x^2+3)}+c.$

 

answered Feb 7, 2013 by sreemathi.v
edited Jul 21, 2013 by vijayalakshmi_ramakrishnans
 
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