■18世紀における微積分(その156)

[5]∫dx/(x^2−1)^1/2=log(x+(x^2−1)^1/2)

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 t=x+(x^2−1)^1/2とおく.

 x=(t^2+1)/2t,dx=(t^2−1)/2t^2dt 

 (x^2−1)^1/2=t−x=(t^2−1)/2t

 したがって,

∫dx/(x^2−1)^1/2=∫2t/(t^2−1)・(t^2−1)/2t^2dt

=∫dt/t=logt=log(x+(x^2−1)^1/2)

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[6]∫dx/(a^2−x^2)=1/2a・∫{1/(a+x)−1/(a−x)}dx=1/2a・log(a+x)/(a−x)

a=1のとき,

∫dx/(1−x^2)=1/2・∫{1/(1+x)−1/(1−x)}dx=1/2・log(1+x)/(1−x)

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