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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function $\int e^{3\log x}(x^4+1)^{-1}$

$\begin{array}{1 1} \large \frac{\log|x^4+1|}{4} +c \\ \large \frac{1}{x^4+1} +c \\ \large \frac{x^4+1}{4} +c \\ \large \frac{1}{\log |x^4+1|} +c \end{array} $

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1 Answer

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Toolbox:
  • (i)$ e^{\large \log x}=x$
  • (ii)If an integral function $f(x)=t,$ then $f'(x)dx=dt$. Hence $\int f(x)dx=\int tdt$
  • (iii)$\int \frac{1}{x}dx=log |x|+c$
Step 1:
Given $I=\int e^{3logx}(x^4+1)^{-1}$
Consider $ e^{3\log x}(x^4+1)^{-1}=e^{\log x^3}(x^4+1)^{-1}$
$\Rightarrow\frac{x^3}{(x^4+1)}$
Therefore $I=\int \frac {x^3dx}{x^4+1}$
Let $x^4=t$
on differentiating w.r.t x we get
$4x^3dx=dt \qquad \Rightarrow x^3dx=\large\frac{dt}{4}$
Substituting for t and dt,
Therefore $I=\frac{1}{4}\int \frac{dt}{t}$
Step 2:
On integrating we get,
$I=\frac{1}{4} \log|t|+c$
Substituting for t we get
$I=\frac{1}{4} \log|x^4+1|+c$
answered Sep 10, 2013 by sreemathi.v
 
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