Question

求不定积分急。。。∫1/√(x²+a²)x²dx

Answer 1

简单步骤如下: ∫ 1/[x²√(x²+a²)] dx 令x=a*tany,dx=a*sec²y dy =a∫ sec²y/[(a²*tan²y)*√(a²*tan²y+a²)] dy =a∫ sec²y/[(a²*tan²y)*√a²sec²y] dy =a/a³*∫ sec²y/(tan²y*secy) dy =1/a²*∫ secy/tan²y dy =1/a²*∫ 1/cosx*cos²y/sin²y dy =1/a²*∫ cosy/sin²y dy =1/a²*∫ cscycoty dy =1/a²*(-cscy)+C =-1/a²*√(x²+a²)/x+C =-√(x²+a²)/(a²x)+C