□投稿者/ 白拓 大御所(607回)-(2006/12/12(Tue) 06:26:52)
| t=√(x^2+1)+xと置換すると dx=√(x^2+1)dt/t 1/t=√(x^2+1)-x (t+1/t)/2=√(x^2+1)
∫{1/√(x^2+1)^3}dx=∫{1/(t{(t+1/t)/2}^2)}dt =4∫{t/(t^2+1)^2}dt=-2/(t^2+1)+C'=-2/({√(x^2+1)+x}^2+1)+C' =-1/{√(x^2+1)*( √(x^2+1)+x )}+C'=-{√(x^2+1)-x}/{√(x^2+1)}+C' =x/√(x^2+1)-1+C'=x/√(x^2+1)+C //
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