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October 11th, 2014, 11:26 AM   #21
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$\displaystyle \int\sqrt{1+\frac{1}{x}}\,dx$

$\displaystyle =\int\frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x}}}\,d x$

$\displaystyle =\int\frac{x+1}{\sqrt{x^2+x}}\,dx$

$\displaystyle =\int\frac{x+1}{\sqrt{\left(x+\frac12\right)^2-\frac14}}\,dx$

$\displaystyle \frac12\cosh(u)=x+\frac12,u=\cosh^{-1}(2x+1)$

$\displaystyle \frac12\sinh(u)\,du=\,dx$

$\displaystyle \int\frac12\cosh(u)+\frac12\,du$

$\displaystyle =\frac12\sinh(u)+\frac12u+C$

$\displaystyle \Leftrightarrow\sqrt{x^2+x}+\frac12\cosh^{-1}(2x+1)+C$
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