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

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Toolbox:
  • (i)If f(x)=t, then f'(x)=dt.Henc $int f(x)dx=\int t.dt$
  • (ii)$\int x^n dx=\frac{x^{n+1}}{n+1}+c$
Given $I=\int\frac{1}{x^2(x^4+1)^\frac{3}{4}}dx$
 
Multiply and divide by $x^{-3}$
 
$I=\large\int\frac{x^{-3}}{x^2x^{-3}(x^4+1)^\frac{3}{4}}dx$
 
$=\large\int\frac{(x^{-3})(x^4+1)^\frac{3}{4}}{x^2(x^4)^{\frac{-3}{4}}}dx$
 
$=\int \frac{1}{x^5}.\bigg(\frac{x^4+1}{x^4}\bigg)^{\frac{-3}{4}}dx$
 
$I=\int \frac{1}{x^5}.\bigg(1+\frac{1}{x^4}\bigg)^{\frac{-3}{4}}dx$
 
Put $\frac{1}{x^4}=t$ on differentiating w.r.t x we get,
 
$\frac{-4}{x^5}dx=dt=>\frac{1}{x^5}dx=\frac{-dt}{4}$
 
On substituting t and dt we get,
 
$I=\frac{-1}{4}\int (1+t)^{\frac{-3}{4}}dx$
 
On integrating we get
 
$-\frac {1}{4}[\frac{1+t}{1/4}]^{1/4}+c$
 
Substituting for t,
 
$I=-\bigg[1+\large\frac{1}{x^4}\bigg]^{\frac{1}{4}}+c$

 

 

answered Feb 15, 2013 by meena.p
 
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