Question

∫dx/(a^2-x^2)

求他的不定积分

Answer 1

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

Answer 2

∫ 1/(a² - x²) dx
= [1/(2a)]∫ [(a - x) + (a + x)]/[(a + x)(a - x)] dx
= [1/(2a)]∫ [1/(a + x) + 1/(a - x)] dx
= [1/(2a)][ln|a + x| - ln|a - x|] + C
= [1/(2a)]ln|(a + x)/(a - x)| + C