Question

求高手解答下∫根号下（x^2+a^2)dx 麻烦写一下详细过程

Answer 1

∫√(x²+a²) dx，x=a*tany，dx=a*sec²y dy
sinx=x/√(x²+a²)，cosx=a/√(x²+a²)
原式= ∫a*secy*a*sec²y dy = a²∫sec³y dy
= a²∫secy dtany
= a²*secy*tany - a²∫tany dsecy
= a²*secy*tany - a²∫tan²y*secy dy
= a²*secy*tany - a²∫(sec²y-1)*secy dy
= a²*secy*tany - a²∫sec³y dy + a²∫secy dy
2a²∫sec³y dy = a²*secy*tany + a²ln|secy+tany|
a²∫sec³y dy = (a²/2)*secy*tany + (a²/2)ln|secy+tany| + C
= (a²/2)(x/a)(1/a)√(x²+a²) + (a²/2)ln|(1/a)√(x²+a²)+(x/a)| + C
= (x/2)√(x²+a²) + (a²/2)ln|x+√(x²+a²)| + C''

追问

怎么看不懂啊，原式第二步到第三步怎么来的