منتديات الرياضيات العربية - عرض مشاركة واحدة

28-06-2008, 12:41 PM

$x = 2\tan u \Rightarrow dx = 2\sec ^2 udu$

$\int {\frac{{dx}}{{(4 + x^2 )^{\frac{3}{2}} }}}= \int {\frac{{2\sec ^2 udu}}{{(4 + 4\tan ^2 u)^{\frac{3}{2}} }} = } \int {\frac{{2\sec ^2 udu}}{{\sqrt {(4 + 4\tan ^2 u)^3 } }}}$

$\int {\frac{{2\sec ^2 udu}}{{8\sqrt {(1 + 1\tan ^2 u)^3 } }}}= \int {\frac{{2\sec ^2 udu}}{{8\sqrt {(\sec ^2 u)^3 } }}}= \int {\frac{{2\sec ^2 udu}}{{8\sec ^3 udu}}}$

$= \int {\frac{1}{{4\sec u}}} du = \frac{1}{4}\int {\cos udu}= \frac{1}{4}\sin u$

$= \frac{1}{4}\frac{x}{{\sqrt {x^2+ 4} }} + c$