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2014-04-23T13:34:29-03:00
 \int\limits { (\sqrt{x} +1)*(x- \sqrt{x} +1)} \, dx

podemos reescrever a  raíz como 
 \sqrt{x} =x^ \frac{1}{2}

reescrevendo a equação
(x^ \frac{1}{2} +1)*(x-x^ \frac{1}{2} +1)

fazendo a multiplicação

(x^ \frac{1}{2} +1)*(x-x^ \frac{1}{2} +1)\\\\\\\\(x^ \frac{1}{2}*x)+(x^ \frac{1}{2}*-x^ \frac{1}{2})+(x^ \frac{1}{2}*1)+(1*x)+(1*-x^ \frac{1}{2})+(1*1)

multiplicação de potencias de mesma base...mante-se a base e soma os expoentes


(x^ \frac{1}{2}*x)+(x^ \frac{1}{2}*-x^ \frac{1}{2})+(x^ \frac{1}{2}*1)+(1*x)+(1*-x^ \frac{1}{2})+(1*1)\\\\\\(x^ \frac{1}{2}^{+1})+(-x^{ \frac{1}{2}^*{2}})+(x^\frac{1}{2} )+(x)+(-x^ \frac{1}{2} )+(1)\\\\(x^ \frac{3}{2} )+(-x^ \frac{2}{2})+(x^ \frac{1}{2})+(x)+(-x^ \frac{1}{2})+1\\\\(x^ \frac{3}{2} )+(-x})+(x^ \frac{1}{2})+(x)+(-x^ \frac{1}{2})+1=\boxed{x^ \frac{3}{2} +1}

(cansei kk) ..

agora substituindo na integral

 \int\limits {x^ \frac{3}{2} +1}} \,. dx

integrando

x^ \frac{3}{2} +1\\\\ \frac{x^ \frac{3}{2}^{+1}  }{\frac{3}{2}^{+1} } +1x\\\\\frac{x^ \frac{5}{2}  }{\frac{5}{2} } +x\\\\ \frac{2x^ \frac{5}{2} }{5} +x\\\\ \frac{2 \sqrt{x^5} }{5} +x+k

K = constante 

\boxed{\boxed{ \int\limits { (\sqrt{x} +1)*(x- \sqrt{x} +1)} \,. dx }}= \boxed{\boxed{ { \frac{2 \sqrt{x^5} }{5} +x+k}  }}

2 5 2
Obrigada, entendi. Eu que me apavorei aqui kkkkkkk. Fácil! (: valeu mesmo.
de nada ;p