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

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Toolbox:
  • $\;$Form of the rational function\[\frac{px+q}{(x+a)(x^2-b^2)}\]
  • $\;$Form of the partial function\[\frac{A}{(x+a)}+\frac{B}{(x+b)}+\frac{C}{(x-b)}\]
Given $I=\int\frac{1}{x(x^4-1)}.$
 
$\frac{1}{x(x^4-1)}=\frac{1}{x(x^2-1)(x^2+1)}.$
 
Resolving into partial function
 
$\;\;\;=\frac{1}{x(x+1)(x-1)(x^2+1)}.$
 
Hence $\frac{1}{x(x^4-1)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1}+\frac{Dx+E}{x^2+1}.$
 
$\Rightarrow 1=A(x+1)(x-1)(x^2+1)+B(x)(x-1)(x^2+1)+C(x)(x^2+1)(x+1)+(Dx+E)(x)(x+1)(x-1).$
 
$1=A(x^4-1)+B(x^4+x^2-x^3-x)+C(x^4+x^2+x^3+x)+(Dx+E)(x^3-x).$
 
Equating the coefficients of $x^4$.
 
0=A+B+C+D----(1)
 
Equating the coefficients of $x^3$.
 
0=-B+C+E-----(2)
 
Equating the coefficients of $x^2$.
 
0=B+C-D-----(3)
 
Equating the coefficients of $x$.
 
0=-B+C=E-----(4)
 
Equating the constant terms
 
1=-A
 
Hence A=-1.
 
substitute this value in equ(1)
 
-1+B+C+D=0.
 
$\Rightarrow B+C+D=1.$-----(5)
 
Add equ(3) and equ(5)
B+C+D=1.
B+C-D=0
________________
2B+2C=1------(6)
Add equ(2) and equ(4)
-B+C+E=0
-B+C-E=0
________________
-2B+2C=0-----(7)
Add equ(6) and equ(7)
2B+2C=1
-2B+2C=0
_______________
4C=1$\Rightarrow C=1/4$
Substitute for B and C in equ (3)
 
B+C-D=0.
 
$\frac{1}{4}+\frac{1}{4}-D=0\Rightarrow D=\frac{1}{2}$.
 
Now substitute for A,B,C in equ(2)
 
-B+C+E=0.
 
$\frac{-1}{4}+\frac{1}{4}+E=0\Rightarrow E=0.$
 
Hence A=-1,B=1/4,C=1/4,D=1/2,E=0.
 
Substituting these values we get,
 
$\frac{1}{x(x^4-1)}=\frac{-1}{x}+\frac{1}{4(x+1)}+\frac{1}{4(x-1)}+\frac{1/2x+0}{2(x^2+1)}.$
 
For integration we can follow the substitution method
 
$\int\frac{1}{x(x^4-1)}=-\int\frac{1}{x}dx+\frac{1}{4}\int{dx}{x+1}+\frac{1}{4}\frac{dx}{x-1}+\frac{1}{2}\int\frac{x}{x^2+1}.$
 
put $x^2+1=t.$ on differentiating we get
 
2xdx=dt$\Rightarrow xdx=dt/2.$
 
On substituting t and dt/2 in the place of xdx and $x^2+1$,
 
$\frac{1}{x(x^4-1)}=\frac{-1}{x}dx+\frac{1}{4}\int\frac{dx}{x+1}+\frac{1}{4}\int\frac{dx}{x-1}+\frac{1}{2}\frac{1}{2}\int\frac{dt}{t}.$
 
On integrating we get,
 
$-log x+\frac{1}{4}log|(x+1)|+\frac{1}{4}log|x-1|+\frac{1}{4}log|t|+c.$
 
$\;\;\;=\frac{1}{4}log\mid\frac{(x+1)(x-1)(t)}{x^4}\mid.$
 
Substituting for t we get
 
$\;\;\;=\frac{1}{4}log\mid\frac{(x^2-1)(x^2+1)}{|x^4|}\mid=\frac{1}{4}log\mid\frac{x^4-1}{x^4}\mid+c.$

 

answered Feb 7, 2013 by sreemathi.v
 
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