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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Evaluate :$\int \large \frac{dx}{x(x^5+3)}$

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Given $I=\large \int\frac{dx}{x(x^5+3)}$
$\textbf{Step 1}$:
Multiply and divide by $x^4$ $\Rightarrow I=\large \int\frac{x^4dx}{x^5(x^5+3)}$
Let $x^5+3=t\rightarrow $Differentiating, $5x^4=dt$ $\rightarrow x^4dx=\large \frac{dt}{5}$
Substituting $x^4dx=\large \frac{dt}{5}$ and $x^5+3= t\;$ we get $\;I=\large \int \frac{dt}{5} \frac{1}{(t-3)t}$
$\textbf{Step 2}$:
Consider $\large \frac{1}{(t-3)t}$. This can be expressed in the form $\large \frac{A}{t}+\frac{B}{t-3}$
$\Rightarrow$ $\large \frac{1}{(t-3)t}$ $= \large \frac{A}{t}+\frac{B}{t-3} = \frac{A(t-3) + Bt}{(t-3)t}$
$\Rightarrow A(t-3)+Bt=1$, with $t = 0$ or $t -3 = 0 \rightarrow t = 3$
$\Rightarrow t=3 \rightarrow 3B=1 \rightarrow$ $B=\large \frac{1}{3}$
$\Rightarrow t=0 \rightarrow -3A=1 \rightarrow A=\large \frac{-1}{3}$
$\Rightarrow \large \frac{1}{(t-3)t}$ $=\large \frac{-1}{3}. \frac{1}{t}$$ +\large \frac{1}{3}. \frac{1}{t-3}$ $= \large \frac{1}{3} (\frac{-1}{t}$$+\large \frac{1}{t-3})$
$\textbf{Step 3}$:
$\Rightarrow I=\large \int \frac{dt}{5} \frac{1}{(t-3)t}$ $=\large \int \frac{1}{5} \large \frac{1}{3} (\frac{-1}{t}$$+\large \frac{1}{t-3})$$\;dt$
$\Rightarrow I = \large \frac{1}{15} \int \frac{-dt}{t}$$+\large \int \frac{dt}{t-3}$
$\Rightarrow I = \large \frac{1}{15}$$ (-\log |t| + \log|t-3|) + c$
$\Rightarrow I = \large \frac{1}{15}$$ \large \frac{|t-3|}{|t|}$$ +c$
Substituting for $\;t = x^5+3$, we get:
$\Rightarrow I = \large \frac{1}{15}$ $\log\large \frac{|x^5+3-3|}{|x^5+3|}$$+c$
$\Rightarrow I =\large \frac{1}{15}$$ \log\large\frac{|x^5|}{|x^5+3|}$$+c$
answered Mar 22, 2013 by balaji.thirumalai
 

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