Edited by Mr0, 10-06-2012 - 16:55.
$\int_{1}^{2}\frac{\sqrt{2+x}}{x(\sqrt{2+x}-\sqrt{2-x})}dx$
Started By Mr0, 10-06-2012 - 16:51
#1
Posted 10-06-2012 - 16:51
$\int_{1}^{2}\frac{\sqrt{2+x}}{x(\sqrt{2+x}-\sqrt{2-x})}dx$
#2
Posted 10-06-2012 - 16:57
$\int_{1}^{2}\frac{\sqrt{2+x}}{x(\sqrt{2+x}-\sqrt{2-x})}$
Tích phân đã cho viết lại:
\[I = \int\limits_1^2 {\frac{{\sqrt {2 + x} \left( {\sqrt {2 + x} + \sqrt {2 - x} } \right)}}{{2{x^2}}}} dx = \frac{1}{2}\left( {\int\limits_1^2 {\frac{{2 + x + \sqrt {4 - {x^2}} }}{{{x^2}}}} } \right)dx\]
\[ = \frac{1}{2}\left[ {\int\limits_1^2 {\left( {\frac{2}{{{x^2}}} + \frac{1}{x} + \frac{{\sqrt {4 - {x^2}} }}{{{x^2}}}} \right)} } \right]dx\]
Việc tính các tích phân thành phần là đơn giản.
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