Question

求不定积分∫x^2*√（x^2+1)dx

Answer 1

x=tant
dx=sec²tdt
原式=∫tan²t·sect·sec²tdt
=∫(sec²t-1)·sec³tdt
=∫sect^5tdt-∫sec³tdt
∫sect^5tdt
=∫sec³tdtant
=sec³ttant-3∫tant·sec²t·tantsectdt
=sec³ttant-3∫sec³t·tan²tdt
=sec³ttant-3∫sec³t·(sec²t-1)dt
=sec³ttant-3∫sec^5tdt+3∫sec³tdt
4∫sec^5tdt=sec³ttant+3∫sec³tdt
∫sec^5tdt=1/4sec³ttant+3/4∫sec³tdt
原式=1/4sec³ttant+3/4∫sec³tdt-∫sec³tdt
=1/4sec³ttant-1/4∫sec³tdt
∫sec³tdt=∫sectdtant
=secttant-∫tant·secttantdt
=sect·tant-∫(sec²t-1)sectdt
两边合并可以解出：∫sec³tdt
下面自己做吧。