高等数学求不定积分，有解析就加悬赏！

Question

如题所述

Answer 1

原式=∫dx/x√[(x+3/2)^2-25/4]
令x+3/2=(5/2)*sect，dx=(5/2)*secttantdt
原式=∫(5/2)*secttantdt/{[(5/2)*sect-(3/2)]*(5/2)*tant}
=∫2sectdt/(5sect-3)
=∫2dt/(5-3cost)
令u=tan(t/2)，则cost=(1-u^2)/(1+u^2)，dt=2du/(1+u^2)
原式=∫[4du/(1+u^2)]/[5-3(1-u^2)/(1+u^2)]
=∫4du/(5+5u^2-3+3u^2)
=∫2du/(1+4u^2)
=arctan(2u)+C
=arctan[2arctan(t/2)]+C
=t/[1-(t/2)^2]+C
=4t/(4-t^2)+C
={4arccos[5/(2x+3)]}/{4-[arccos[5/(2x+3)]]^2}+C，C是任意常数