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Integrate the function $\frac{\large 1}{\large x-\sqrt x}$

$\begin{array}{1 1} 2 \log\;|\sqrt x-1|+c \\ 2 \log\;|\sqrt x+1|+c \\ \log\;|\sqrt x-1|+c \\ \log\;|\sqrt x+1|+c \end{array} $

1 Answer

Toolbox:
  • Method of substitution:
  • Given $\int f(x)dx$ can be transformed into another form by changing independent variable x to t by substituting x=g(t).
  • Consider $I=\int f(x)dx.$
  • Put x=g(t) so that $\frac{dx}{dx}=g'(t).$
  • dx=g'(t)dt.
  • Thus $ I=\int f(g(t).g'(t))dt.$
Given $ I=\int \frac{1}{x-\sqrt x}dx.=\int\frac{1}{\sqrt x(\sqrt x-1)}dx.$
 
Let $(\sqrt x-1)$ be t .
 
$\frac{1}{2\sqrt x}dx=dt\;\Rightarrow \frac{1}{\sqrt x}dx=2dt$.
 
Now we can Substitute for t and dt in I we get,
 
$\int \frac{1}{\sqrt x(\sqrt x-1)}=\int\frac{2}{t}dt$.
 
 
On integrating we get,
 
$\;\;\;=2log|t|+c$.
 
Substituting back for t we get,
 
$\;\;\;=2log|\sqrt x-1|+c.$
 
Hence $ \int\frac{1}{x-\sqrt x}dx=2log|\sqrt x-1|+c.$

 

answered Jan 28, 2013 by sreemathi.v
edited Sep 20, 2013 by sreemathi.v
 
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