Question

求助一道关于不定积分的题，谢谢!

Answer 1

∫ √(ax²+bc+c) dx
=∫ √[(√ax+b/(2√a))²+(4ac-b²)/(4a)] dx
令u=√ax+b/(2√a),du=√a dx
=1/√a*√[u²+(4ac-b²)/(4a)] du
令t=2u/[√(4ac-b²)/a],dt=2/√[(4ac-b²)/a] du
=c/√a-b²/(4a^3/2)*∫√(1+t²) dt
令t=tanp,dt=sec²p dp
√(1+t²)=secp
=c/√a-b²/(4a^3/2)-∫sec³p dp
=c/√a-b²/(4a^3/2)-(1/2)[secp*tanp+ln|secp+tanp|]+C
逐个代回下面3个,
p=arctan(t)
t=2u/√[(4ac-b²)/a]
u=√ax+b/(2√a)
最后得出答案为：
1/(8a^3/2)*{2√a*(2ax+b)√(ax²+bx+c)-(b²-4ac)ln[2√a*√(ax²+bx+c)+2ax+b]}+C