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Evaluate the following: $\int\frac{x^2dx}{x^4-x^2-12}$

$\begin{array}{1 1}(A)\;\large\frac{1}{7}\log \large\frac{|x-2|}{|x+2|}-\frac{\sqrt 3}{7} \tan ^{-1} \bigg(\large\frac{x}{\sqrt 3}\bigg)+c \\ (B)\;\large\frac{1}{7} \log \large\frac{|x-2|}{|x+2|}+\frac{\sqrt 3}{7} \tan ^{-1} \bigg(\large\frac{x}{\sqrt 3}\bigg)+c \\ (C)\;\large\frac{1}{7} \log \large\frac{|x-2|}{|x+2|}+\frac{ 3}{\sqrt 7} \tan ^{-1} \bigg(\large\frac{x}{\sqrt 3}\bigg)+c \\ (D)\;\large\frac{1}{7} \log \large\frac{|x-2|}{|x+2|}-\frac{3}{\sqrt 7} \tan ^{-1} \bigg(\large\frac{x}{\sqrt 3}\bigg)+c\end{array} $

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  • If a rational expression, whose degree of the numerator is greater then or equal to that of the denominator, it is said to be improper.
  • A rational expression of the from $\large\frac{x}{(x+a)(x-a)}$ can be resolved as $\large\frac{A}{(x+a)}+\frac{B}{(x-a)}$
Here $x^4-x^2-12$ can be factorized as $(x^2-4)(x^2+3)$
Hence $\large\frac{x^2}{x^4-x^2-12}$ can be resolved as
To evaluate the values for A,B,C and D put x=2
Therefore $ 4=28 B \qquad B=\frac{1}{7}$
Put $x=-2$
Therefore $ 4=-28A=> \qquad A=\large\frac{-1}{7}$
Put $x=0$
$=> -6A+6B-4D=0$
$-6 \times \large\frac{-1}{7} $$+ 6 \times \large\frac{1}{7}-$$4D=0$
Put $x=1$
$=>1=-4 \times \bigg(\large\frac{-1}{7}\bigg)$$+12 \bigg(\large\frac{1}{7}\bigg)-3C$
$(ie) \large\frac{4}{7}+\frac{12}{7}-$$3C=1$
$=>3C=\large\frac{9}{7}$ therefore $C=\large\frac{3}{7}$
Hence $A=-1/7,B=1/7\; and \;c=3/7$
Now substituting the value for A,B and C we get
Hence $I=\large\frac{-1}{7} \int \frac{1}{(x+2)}dx+\frac{1}{7} \int \frac{dx}{(x-2)}+\frac{3}{7}\int \frac{dx}{x^2+3}$
On integrating we get
$I=-\large\frac{1}{7}$$\log |x+2|+\large\frac{1}{7} $$\log|x-2|+\large\frac{3}{7} \times \frac{1}{\sqrt 3} $$\tan ^{-1} (\large\frac {x}{\sqrt 3})$
$=\large\frac{1}{7} $$\log \large\frac{|x-2|}{|x+2|}+\frac{\sqrt 3}{7} $$\tan ^{-1} \bigg(\large\frac{x}{\sqrt 3}\bigg)+c$
answered Apr 21, 2013 by meena.p
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