17/09/17 07:35:47.61 aM377AWr.net
>>309
不定積分だが
∫ dx / (x^2 + 1)
= ∫ dx / (i + x)(i - x)
= (1/2i) ∫ (1/(i + x) + 1/(i - x)) dx
= (1/2i) (log(i + x) - log(i - x)) + C
= (1/2i) log((i + x)/(i - x)) + C
= y + C
とおくと
(i + x)/(i - x) = exp(2iy)
x = -i(exp(2iy) - 1) / (exp(2iy) + 1)
x = -i(exp(iy) - exp(-iy)) / (exp(iy) + exp(-iy))
x = -i (2i sin y)/(2 cos y) = tan y
ゆえに y = arctan x