Question

求∫1/[(a^2+x^2)^2]dx的不定积分

Answer 1

∫ dx/(a² + x²)

= ∫ dx/[a²(1 + x²/a²)]

= (1/a²)∫ dx/(1 + x²/a²)

= (1/a²)∫ d(x/a · a)/(1 + x²/a²)

= (1/a²)(a)∫ d(x/a)/(1 + x²/a²)

= (1/a)∫ d(x/a)/[1 + (x/a)²]

= (1/a)arctan(x/a) + C