My Math Forum Plz, an integration ! important in my exam tomo.

 Calculus Calculus Math Forum

 October 11th, 2014, 11:26 AM #21 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond $\displaystyle \int\sqrt{1+\frac{1}{x}}\,dx$ $\displaystyle =\int\frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x}}}\,d x$ $\displaystyle =\int\frac{x+1}{\sqrt{x^2+x}}\,dx$ $\displaystyle =\int\frac{x+1}{\sqrt{\left(x+\frac12\right)^2-\frac14}}\,dx$ $\displaystyle \frac12\cosh(u)=x+\frac12,u=\cosh^{-1}(2x+1)$ $\displaystyle \frac12\sinh(u)\,du=\,dx$ $\displaystyle \int\frac12\cosh(u)+\frac12\,du$ $\displaystyle =\frac12\sinh(u)+\frac12u+C$ $\displaystyle \Leftrightarrow\sqrt{x^2+x}+\frac12\cosh^{-1}(2x+1)+C$

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