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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the rational functions\[\frac{1}{x(x^n+1)}\] \[Hint:multiply\;numerator\;and\;denominator\;by\;x^{n-1}and\;put\;x^n=t\]

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Toolbox:
  • $(i)\;$Form of the ration function\[\frac{px+q}{(x+a)(x+b)}\]
  • $\;$Form of the partial function\[\frac{A}{(x+a)}+\frac{B}{(x+b)}\]
  • $(ii)\;\int\frac{dx}{(x+a)}=log|x+a|+c.$
Given:$I=\int\frac{1}{x(x^n+1}dx.$
 
Multiply the numerator and denominator by $x^n-1$.
 
then,$I=\int\frac{x^{n-1}}{x(x^n+1)(x^{n-1}}dx.$
 
But $x'(x^{n-1})=x^n.$
 
Therefore $I=\int\frac{x^{n-1}}{x^n(x^n+1)}dx.$
 
Now put $x^n=t,$ then $nx^{n-1}dx=dt.$
 
$\Rightarrow x^{n-1}dx=\frac{dt}{n}.$
 
Substituting this we get,
 
$I=\int\frac{dt/n}{t(t+1)}=\frac{1}{n}\int\frac{dt}{t(t+1)}$.
 
Resolving into partial fraction
 
Let $\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{t+1}.$
 
$\Rightarrow 1=A(t+1)+Bt.$
 
Now equating the coefficient of t we get,
 
0=A+B-------(1)
 
Now equating the constant term,
 
1=A,Hence A=1.
 
Substituting for A in equ (1)
 
1+B=0$\Rightarrow B=-1.$Hence A=1 and B=-1.$
 
Now substituting for A and B,we get
 
$\frac{1}{t(t+1)}=\frac{1}{t}-\frac{1}{t+1}.$
 
Therefore $I=\frac{1}{n}\begin{bmatrix}\int\frac{1}{t}dt-\int\frac{1}{t+1}dt\end{bmatrix}$
 
On integrating we get,
 
$\frac{1}{n}log|t|-\frac{1}{n}log|t+1|+C.$
 
Substituting for t we get,
 
$\;\;\;=\frac{1}{n}log|x^n|-\frac{1}{n}log|x^n+1|+C.$
 
$log a-log b=log\mid\frac{a}{b}\mid$,similarly
 
$\;\;\;=\frac{1}{n}log\mid\frac{x^n}{x^n+1}\mid+c.$

 

answered Feb 7, 2013 by sreemathi.v
 
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