Question

∫[1/(x^2-a^2)]dx

要过程，谢谢！

Answer 1

∫[1/(x^2-a^2)]dx
=∫[1/(x+a)(x-a)]dx
=∫[1/(x-a)-1/(x+a)]/2a dx，拆项
=(1/2a)∫[1/(x-a)-1/(x+a)]dx
=(1/2a)∫[1/(x-a)]dx-(1/2a)∫[1/(x+a)]dx
=(1/2a)ln|x-a|-(1/2a)ln|x+a|+C,C为常数
=(1/2a) ln|(x-a)/(x+a)| + C

Answer 2

∫[1/(x^2-a^2)]dx
=2/a*∫[1/(x-a)-1/(x+a)]dx
=2/a*{∫[1/(x-a)]dx-∫[1/(x+a)]dx}
=2/a*[ln(x-a)-ln(x+a)]
=2/a*ln[(x-a)/(x+a)]

Answer 3

1/(x^2-a^2)=1/(2a)*(1/(x-a)-1/(x+a))
∫[1/(x^2-a^2)]dx
=1/(2a)*(∫[1/(x-a)]dx-∫[1/(x+a)]dx)
=1/(2a)*(ln|x-a|-ln|x+a|)+C
=1/(2a)*(ln|(x-a)/(x+a)|+C