Question

求(arctanx)^(1/2)/(1+x^2)的不定积分

Answer 1

∫ arctanx / (1+x²)^(3/2) dx

= ∫ arctanx d[x/√(x²+1)],分部积分法,∫ dx/(1+x²)^(3/2) = x/√(x²+1)

= [x/√(x²+1)]arctanx - ∫ x/√(x²+1) d(arctanx),(arcanx)' = 1/(x²+1)

= x*arctanx / √(x²+1) - ∫ x/(x²+1)^(3/2) dx

= x*arctanx / √(x²+1) - (1/2)∫ d(x²+1)/(x²+1)^(3/2)

= x*arctanx / √(x²+1) - (1/2)*(x²+1)^(-3/2+1) / (-3/2+1) + C

= x*arctanx / √(x²+1) - (1/2)(-2)(x²+1)^(-1/2) + C

= x*arctanx / √(x²+1) + 1/√(x²+1) + C

= (x*arctanx + 1) / √(x² + 1) + C