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

$(a)\; 3\log|x+2|+\large \frac{7}{(x+2)}+c\qquad(b)\;3\log|x+2|-\large \frac{7}{(x+2)}+c\qquad(c)\; \log|x+2|+\large \frac{7}{(3x+1)}+c\qquad(d)\;3\log|x+2|+\large \frac{7}{(3x-1)}+c$

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Toolbox:
  • $(i)\;$Form of the rational function\[\frac{px+q}{(x+a)^2}\]
  • $\;$Form of the partial function\[\frac{A}{(x+a)}+\frac{B}{(x+a)^2}\]
  • $(ii)\;\int\frac{dx}{(x+a)}=log|x+a|+c.$
  • $(iii)\;\int\frac{dx}{(x+a)^2}=\frac{-1}{(x+a)}$
Given:$I=\int\frac{3x-1}{(x+2)^2}dx.$
Let $\frac{3x-1}{(x+2)^2}=\frac{A}{(x+2)}+\frac{B}{(x+2)^2}.$
$\Rightarrow 3x-1=A(x+2)+B$
Equating the coefficients of x,
3=A$\Rightarrow A=3.$
Equating the constant terms,
-1=2A+B
since A=3.
-1=2(3)+B.
$\Rightarrow B=-1-6=7.$
Hence A=3 and B=7.
Hence $\frac{3x-1}{(x+2)^2}=\frac{3}{(x+2)}-\frac{7}{(x+2)^2}.$
Substitute for A in equ(1)
$I=\int\frac{3}{(x+2)}dx-7\int\frac{dx}{(x+2)^2}.$
On integrating we get,
$\;\;\;=3log|x+2|-7\frac{(x+2)^{-2+1}}{-2+1}.$
$\;\;\;=3log|x+2|-7\frac{-1}{(x+2)}+c.$
$\;\;\;=3log|x+2|+\frac{7}{(x+2)}+c.$
answered Jan 9, 2014 by yamini.v
 
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