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

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

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 $\int\frac{1}{x-\sqrt x}dx=2log|\sqrt x-1|+c.$

edited Sep 20, 2013