\[I = \int {\frac{{lnx}}{{x*\sqrt {1 - 4sinx - {{(lnx)}^2}} }}} \]
(Đề thi cuối kỳ đh Dược Hà Nội.)
Edited by Nguyen Lam Thinh, 06-12-2012 - 19:24.
$I = \int {\frac{{lnx}}{{x*\sqrt {1 - 4sinx - {{(lnx)}^2}} }}}$
Started By Dont Cry, 06-12-2012 - 15:48
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