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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the integral using substitution\[\int\limits_0^2\frac{dx}{x+4-x^2}\]

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Toolbox:
  • (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
  • $\int \frac{dx}{a^2-x^2}=\frac{1}{2a}log \frac{|a+x|}{|a-x|}+c$
Given $\int\limits_0^2\frac{dx}{x+4-x^2}$
 
Consider $ -x^2+x+4$
 
$=-(x^2-x-4)$
 
This can be written as $-\bigg[(x-1/2)^2-\frac{1}{4}-4\bigg]$
 
$=-\bigg[(x-1/2)^2-\frac{17}{4}\bigg]$
 
Therefore $I=\int \limits_0^2 \frac{dx}{(\frac{\sqrt {17}}{2})^2-{(x-1/2)^2}}$
 
This is of the form $ \int \large\frac{dx}{a^2-x^2}=\frac{1}{2a} log \frac{|a+x|}{|a-x|}+c$
 
In the place of x we have (x-1/2) and in the place of a we have $\frac{\sqrt {17}}{2}$
 
Hence on integrating,
 
Therefore $I=\bigg[\large\frac{1}{2.\frac{\sqrt {17}}{2}} log \frac{\bigg|\frac{\sqrt {17}}{2}+(x-1/2)\bigg|}{\bigg|\frac{\sqrt {17}}{2}-(x-
1/2)\bigg|}\bigg]_0^2$
 
$=\bigg[\large\frac{1}{\sqrt {17}} log \frac{\bigg|\frac{\sqrt {17}}{2}+\frac{(2x-1)}{2}\bigg|}{\bigg|\frac{\sqrt {17}}{2}-\frac{(2x-1)}{2}\bigg|}\bigg]_0^2$
 
Applying limits and simplifying we get,
 
$I=\bigg[\large\frac{1}{\sqrt {17}} log \frac{|\sqrt {17}+3|}{|\sqrt {17}-3|} \times \frac{|\sqrt {17}+1|}{|\sqrt {17}-1|} \bigg]$
 
$=\large\frac{1}{\sqrt {17}} log \bigg[\frac{17+3+4\sqrt {17}}{17+3-4\sqrt {17}}\bigg]$
 
$=\large\frac{1}{\sqrt {17}} log \bigg[\frac{20+4\sqrt {17}}{20-4\sqrt {17}}\bigg]$
 
$=\large\frac{1}{\sqrt {17}} log \bigg[\frac{5+\sqrt {17}}{5-\sqrt {17}}\bigg]$
 
Multiplying and dividing by the conjugat $5+\sqrt {17}$
 
$I=\large\frac{1}{\sqrt {17}} log \bigg[\frac{5+\sqrt {17}}{5-\sqrt {17}} \times \frac{5+\sqrt {17}}{5+\sqrt {17}} \bigg]$
 
$=\large\frac{1}{\sqrt {17}} log \bigg[\frac{25+17+10\sqrt {17}}{8}\bigg]$
 
$=\large\frac{1}{\sqrt {17}} log \bigg[\frac{42+10\sqrt {17}}{8}\bigg]$
 
$I=\large\frac{1}{\sqrt {17}} log \bigg[\frac{21+5\sqrt {17}}{8}\bigg]$

 

 

answered Feb 12, 2013 by meena.p
 
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