منتديات الرياضيات العربية - عرض مشاركة واحدة

17-06-2008, 08:18 PM

$\int {\sqrt {e^x- 1} dx}$

${\rm{u = }}\sqrt {e^x- 1}\Rightarrow u^2= e^x- 1 \Rightarrow e^x= u^2+ 1$
$2udu = e^x dx \Rightarrow dx = \frac{{2udu}}{{e^x }} = \frac{{2udu}}{{u^2+ 1}}$
$\int {\sqrt {e^x- 1} dx}= \int {u\frac{{2udu}}{{u^2+ 1}}}$

$= \int {\frac{{2u^2 du}}{{u^2+ 1}}}= \int {2du - \int {\frac{2}{{u^2+ 1}}du} }$

$= 2u - 2\tan ^{ - 1} (u)$
when x=0 u=0
${\rm{when x = ln2u = }}\sqrt {e^{\ln 2}- 1}= \sqrt {2 - 1}= 1$
=the integral

$\left. {2u - 2\tan ^{ - 1} (u)} \right|_0^1= (2 - 2\frac{\pi }{4}) - (0)$
$= 2(1 - \frac{\pi }{4})$