 | 2007/05/01(Tue) 16:36:31 編集(投稿者)
(与式)=∫dx/[a{x+b/(2a)}^2+c-(b^2)/(4a)]
=∫dx/[a{x+b/(2a)}^2+(4ac-b^2)/(4a)]
={4a/(4ac-b^2)}∫dx/[1+{4a^2/(4ac-b^2)}{x+b/(2a)}^2]
ここで
D=b^2-4ac<0、a>0ゆえ
4a^2/(4ac-b^2)>0
よって
(与式)={4a/(4ac-b^2)}・{1/√{4a^2/(4ac-b^2)}}
・arctan[√{4a^2/(4ac-b^2)}{x+b/(2a)}]+C
=(1/a)√{4a^2/(4ac-b^2)}arctan[√{4a^2/(4ac-b^2)}{x+b/(2a)}]+C
={2/√(4ac-b^2)}arctan{(2ax+b)/√(4ac-b^2)}+C
(C:積分定数)
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