منتديات الرياضيات العربية - عرض مشاركة واحدة

05-06-2008, 09:04 AM

حل آخر
let

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

$\int {\sqrt {1 + u} } (2)udu = \int {2u} \sqrt {1 + u} du$
let

$z = 1 + u \Rightarrow dz = du$

$\int {\sqrt z } (2)(z - 1)dz = \int {(2z - 2)z^{\frac{1}{2}} } dz$

$= \int {2z^{\frac{3}{2}} }- 2z^{\frac{1}{2}} .dz = 2\frac{{2z^{\frac{5}{2}} }}{5} - 2(2)\frac{{z^{\frac{3}{2}} }}{3}$

$= \frac{4}{5}\sqrt {z^5 }- \frac{4}{3}\sqrt {z^3 }= \frac{4}{5}\sqrt {(1 + u)^5 }- \frac{4}{3}\sqrt {(1 + u)^3 }$

$= \frac{4}{5}\sqrt {(1 + \sqrt x )^5 }- \frac{4}{3}\sqrt {(1 + \sqrt x )^3 }+ c$