I=$\large \int_{0}^{\propto }\frac{arctanx}{(1+x^2)^{\frac{3}{2}}}dx$
I=$\large \int_{0}^{\propto }\frac{arctanx}{(1+x^2)^{\frac{3}{2}}}dx$
Bắt đầu bởi bachkhoa, 03-08-2015 - 23:17
#1
Đã gửi 03-08-2015 - 23:17
#2
Đã gửi 25-08-2015 - 20:58
I=$\large \int_{0}^{\propto }\frac{arctanx}{(1+x^2)^{\frac{3}{2}}}dx$
\[I = \mathop {\lim }\limits_{a \to + \infty } \int\limits_0^a {\sqrt {\arctan x} d\arctan x} = \mathop {\lim }\limits_{a \to + \infty }{\frac{2}{3}{{(\arctan a)}^{3/2}} = \frac{2}{3}{\pi ^{3/2}}} \]
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