K!! Posted January 2, 2008 Share Posted January 2, 2008 I'd like to initiate a kind of Integral Marathon. This is simple, person who solves a problem must receive a confirmation whether the answer is correct or not. In case where the answer is correct, solver may post the next integral. (Of course, indefinite & finite integrals are allowed.) Let's start with an easy one: Solve [math]\int {\frac{1} {{x\sqrt {x^2 - x} }}\,dx}[/math]. Link to comment Share on other sites More sharing options...
Ragib Posted January 3, 2008 Share Posted January 3, 2008 [math]\int \frac{1}{x\sqrt{x}\sqrt{x-1}} dx[/math] Let [math]x=\cosh^2 u[/math], then [math]dx = 2 \cosh u \sinh u du[/math]. [math]\int \frac{2\cosh u \sinh u}{\cosh^3 u \sinh u} du[/math] = 2 \int \sech^2 u du [math]= 2\tanh u + C [/math] [math]= 2\tanh (\cosh^{-1} \sqrt{x} ) + C [/math] EDIT: LaTeX was not working for those two lines, so I ommitted the tags so people could see what I meant to type, and hopefully someone can correct my 'LaTeX syntax error'. (Fixed the LaTeX for you. It turns out there's no \arccosh command in LaTeX, so I had to make do with cosh-1. Oh well. -- Cap'n) Link to comment Share on other sites More sharing options...
K!! Posted January 3, 2008 Author Share Posted January 3, 2008 Okay well [math]\int {\frac{1} {{x\sqrt {x^2 - x} }}\,dx} = \frac{{2\sqrt {x^2 - x} }} {x} + k[/math]. Here's my approach: Set [math]x=\frac1u,[/math] the integral becomes [math]- \int {\frac{1} {{\sqrt {1 - u} }}\,du} = 2\sqrt {1 - u} + k[/math]. The rest follows. -- Post your integral now. Link to comment Share on other sites More sharing options...
Ragib Posted January 3, 2008 Share Posted January 3, 2008 How about [math]\int^1_0 \frac{\ln (1+x)}{x} dx[/math] . Link to comment Share on other sites More sharing options...
K!! Posted January 3, 2008 Author Share Posted January 3, 2008 It is a well known one; its solution is based on series. The answer is [math]\frac{{\pi ^2 }} {{12}}[/math]. Can you edit your post with another integral? Link to comment Share on other sites More sharing options...
K!! Posted February 5, 2008 Author Share Posted February 5, 2008 Okay well, after series expansion we have: [math]\begin{aligned} \int_0^1 {\frac{{\ln (1 + x)}} {x}\,dx} &= \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k }} {{k + 1}}\left\{ {\int_0^1 {x^k \,dx} } \right\}}\\ &= \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k }} {{(k + 1)^2 }}}\\ &= \frac{{\pi ^2 }} {{12}}. \end{aligned}[/math] Link to comment Share on other sites More sharing options...
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