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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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  • (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 $x^{-3}$
$=\int \frac{1}{x^5}.\bigg(\frac{x^4+1}{x^4}\bigg)^{\frac{-3}{4}}dx$
$I=\int \frac{1}{x^5}.\bigg(1+\frac{1}{x^4}\bigg)^{\frac{-3}{4}}dx$
Put $\frac{1}{x^4}=t$ on differentiating w.r.t x we get,
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,



answered Feb 15, 2013 by meena.p
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