Article View: fj.sci.math
Article #13829$BBg3X=iG/5iITDj@QJ,7W;;LdBj<ALd(B
From: KN879682
Date: Sun, 24 Mar 1996 00:00
Date: Sun, 24 Mar 1996 00:00
37 lines
1810 bytes
1810 bytes
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\documentstyle[a4j]{jarticle}
\begin{document}
$\displaystyle \int\frac{dx}{(x^{2}+a^{2})^{3/2}}$ $B$r5a$a$k!%(B
$$(x^{2}+a^{2})^{\frac{1}{2}}=t-x$$
$B$H$*$/$3$H$K$h$j!$(B
$$x=\frac{t^{2}-a^{2}}{2t}\quad ,\quad
dx=\frac{t^{2}+a^{2}}{2t^{2}}dt.$$
$B$^$?$3$l$h$j!$(B
$$(x^{2}+a^{2})^{\frac{1}{2}}=t-x=\frac{t^{2}+a^{2}}{2t}$$
$B$rF@$k!%$7$?$,$C$F!$(B
$$\int\frac{dx}{(x^{2}+a^{2})^{\frac{3}{2}}}=
\int\frac{8t^{3}}{(t^{2}+a^{2})^{3}}\cdot\frac{t^{2}+a^{2}}{2t^{2}}dt
=\int\frac{4t}{(t^{2}+a^{2})^{2}}dt.$$
$B$5$i$K!$$3$3$G(B $\displaystyle t^{2}+a^{2}=u$ $B$H$/$3$H$K$h$C$F!$(B
$\displaystyle 2tdt=du$ $B$H$J$k$+$i>e<0$N:G1&JU$O(B
$\displaystyle \int\frac{2}{u^{2}}du$ $B$H$J$k!%(B
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$$\int\frac{2}{u^{2}}du=-\frac{2}{u}=
-\frac{1}{x^{2}+x\sqrt{x^{2}+a^{2}}+a^{2}}$$
$B$H7W;;$5$l$k!%(B
$B$A$J$_$K@52r$O(B$\displaystyle \frac{x}{a^{2}\sqrt{x^{2}+a^{2}}}$
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\end{document}
--- LaTeX END
$B?@F`@n8)(B $B>.ARK.I'(B
E-mail : KN879682@copernicus.or.jp
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X-Google-Language: Japanese,JIS X-Google-Thread: f4956,18929bc4c88b5c88,start X-Google-Attributes: gidf4956,public From: KN879682@copernicus.or.jp Subject: $BBg3X=iG/5iITDj@QJ,7W;;LdBj<ALd(B Date: 1996/03/24 Message-ID: <960324160709.29420@copernicus.or.jp> X-Deja-AN: 144596311 organization: COPERNICUS, japan newsgroups: fj.sci.math