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Evaluate the definite integrals\[\int\limits_0^1\frac{dx}{\sqrt{1+x}-\sqrt x}\]

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Toolbox:
  • (i)$\int \limits_a^bf(x)dx=F(b)-F(a)+c$
  • $\int x^n dx=[\frac{x^{n+1}}{n+1}]+c$
Given $I=\int\limits_0^1\frac{dx}{\sqrt{1+x}-\sqrt x}$
 
Let us multiply and divide by its conjujate,
 
Therefore $I=\int\limits_0^1\frac{dx}{\sqrt{1+x}-\sqrt x} \times \frac{\sqrt {1+x}+\sqrt x}{\sqrt {1+x}+\sqrt x}$
 
$=\int\limits_0^1\frac{\sqrt {1+x}+\sqrt x}{1+x-x}dx$
 
$=\int\limits_0^1({\sqrt {1+x}+\sqrt x})dx$
 
On seperating the terms,
 
$I=\int\limits_0^1\sqrt {1+x}dx+\int \limits_0^1\sqrt x\;dx$
 
On integrating we get,
 
$\bigg[\frac{(1+x)^{1/2+1}}{1/2+1}\bigg]_0^1+\bigg[\frac{(x)^{1/2+1}}{1/2+1}\bigg]_0^1+c$
 
$=\bigg[\frac{(1+x)^{3/2}}{3/2}\bigg]_0^1+\bigg[\frac{(x)^{3/2}}{3/2}\bigg]_0^1+c$
 
$=\frac{2}{3}[(1+x)^{3/2}]_0^1+\frac{2}{3}[(x)^{3/2}]_0^1+c$
 
On applying limits,
 
$=\frac{2}{3}[(2)^{3/2}-1]+\frac{2}{3}[(1)^{3/2}]+c$
 
$=\frac{2}{3} \bigg\{2^{3/2}-1+1 \bigg\}+c$
 
$I=\frac{2}{3}[2^{\frac{3}{2}}]+c$

 

 

answered Feb 19, 2013 by meena.p
 
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