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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Evaluate : $ \int\large\frac{(3x-2)}{(x+1)^2(x+3)} $$dx$

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Toolbox:
  • Form of the rational function $\large\frac{px+q}{(x^2+a)(x^2+b)}$
  • Form of the partial function $\large\frac{Ax+B}{(x^2+a)}+\frac{Cx+D}{(x^2+b)}$
  • $\int\large\frac{dx}{x}$$=log|x|+c.$
Step 1:
$I=\int\large\frac{(3x-2)}{(x+1)^2(x+3)^2}$$dx$
Consider $\large\frac{3x-2}{(x+1)^2(x+3)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+3}$
$\Rightarrow (3x-2)=A(x+1)(x+3)+B(x+3)+C(x+1)^2$
$\Rightarrow A(x^2+4x+3)+B(x+3)+C(x^2+2x+1)$
Step 2:
Now compare the coefficient of like terms,
For $x^2 \quad A+C=0$
For $x \quad 4A+B+2C=3$
For the constant term $ \quad 3A+3B+C=-2$
Solving the above equations we get,
$A=\large\frac{11}{4}$$B=\large\frac{-5}{2}$ and $C=\large\frac{-11}{4}$
Step 3:
$\large\frac{3x-2}{(x+1)^2(x+3)}=\frac{11}{4(x+1)}-\frac{5}{2(x+1)^2}-\frac{11}{4(x+3)}$
$5=\large\frac{11}{4}\int\frac{dx}{x+1}-\frac{5}{2}\int\frac{dx}{(x+1)^2}-\frac{11}{4}\int\frac{dx}{x+3}$
On integrating we get,
$I=\large\frac{11}{4}$$\log\mid x+1\mid+\large\frac{5}{2(x+1)}-\frac{11}{4}$$\log\mid x+3\mid+c$
$\Rightarrow \large\frac{11}{4}$$\log\bigg|\large\frac{x+1}{x+3}\bigg|+\frac{5}{2(x+1)}+c$
answered Sep 30, 2013 by sreemathi.v
 
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