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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the rational functions\[\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\]

$\begin{array}{1 1}x-\frac{2}{3}\tan^{-1}\big(\frac{x}{\sqrt 3}\big)+3\tan^{-1}\big(\frac{x}{2}\big)+c. \\x+\frac{2}{3}\tan^{-1}\big(\frac{x}{\sqrt 3}\big)-3\tan^{-1}\big(\frac{x}{2}\big)+c. \\ x+\frac{2}{3}\tan^{-1}\big(\frac{x}{\sqrt 3}\big)+3\tan^{-1}\big(\frac{x}{2}\big)+c. \\ \frac{2}{3}\tan^{-1}\big(\frac{x}{\sqrt 3}\big)+3\tan^{-1}\big(\frac{x}{2}\big)+c.\end{array} $

Can you answer this question?
 
 

1 Answer

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Toolbox:
  • If f(x)=t,then on differentiating f'(x)dx=dt.
  • Thus $\int f(x)dx=\int tdt.$
  • (ii)If the rational function is improper in nature.we can divide and separate the terms to make it a proper rational function.
  • (iii)$\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\big(\frac{x}{a}\big)+c.$
  • (iv)$\int\frac{dx}{x+a}=log|x+a|+c.$
Given $I=\int\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}.$
 
Let $x^2=t,on differentiating we get
 
2xdx=dt$\Rightarrow xdx=\frac{dt}{2}.$
 
Therefore $\frac{(x^2+1)(x^2+2)}{x^2+3)(x^2+4)}=\frac{(t+1)(t+2)}{(t+3)(t+4)}.$
 
$\;\;\;\qquad\qquad\qquad\;\;\;=\frac{t^2+3t+2}{t^2+7t+12}.$
 
But this is an improper rational fraction,so let us divide to make it a proper rational function.
 
$\frac{t^2+5t+4}{t^2-2t-15}=1-\frac{4t+10}{t^2+7t+12}$
 
On separating the term
 
$I=\int dx+\int\frac{4t+10}{t^2+7t+12}.\frac{dt}{2}.$
 
Let $\frac{4t+10}{t^2-2t-15}=\frac{7t+19}{(t+3)(t+4)}$
 
Resolving into partial fraction
 
$\;\;\;=\frac{A}{t+4}+\frac{B}{t+3}.$
 
$\Rightarrow 4t+10=A(t+4)+B(t+3).$
 
Equating the coefficients of t we get,
 
4=A+B.-----(1)
 
Equating the constant terms we get
 
10=4A+3B------(2)
 
Multiply equ(1) by 3 and subtract from equ(2)
 
4A+3B=10
3A+3B=12
_________________
A=-2.
Substitute for A in equ(1)
 
B-2=4$\Rightarrow B=6.$
 
Hence A=-2,B=6.
 
On substituting for A and B,
 
$\frac{4t+10}{(t+3)(t+4)}=\frac{6}{t+3}-\frac{2}{t+4}.$
 
Now substituting for t we get
 
$I=\int dx-\int\frac{2}{x^2+3}dt+\int\frac{6}{x^2+2^2}dt.$
 
On integrating we get
 
$I=x-2.\frac{1}{\sqrt 3}\tan^{-1}\big(\frac{x}{\sqrt 3}\big)+6.\frac{1}{2}\tan^{-1}\big(\frac{x}{2}\big)+c.$
 
$\;\;\;=x-\frac{2}{3}\tan^{-1}\big(\frac{x}{\sqrt 3}\big)+3\tan^{-1}\big(\frac{x}{2}\big)+c.$

 

answered Feb 7, 2013 by sreemathi.v
 

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