谁能帮忙一下求∫x²√（x²+a²）dx和∫（2x+1）sin2x dx的解题过程

Question

参考答案：(1/8)a^4arcsin(x/a)-(1/8)x（a²-2x²）√（a²-x²）+C和x/[a²√（x²+a²）]+C

Answer 1

∫x²√（x²+a²）dx
解：令x=atant，则原式=a^4∫sin^2(t)/cos^5(t)dt=a^4[∫sec^5(t)dt+∫sec^3(t)dt]
首先求∫sec^3(t) dt：记I=∫sec^3(t) dt，则I
=∫sec(t)*sec^2(t) dt
=∫sec(t)*[tan(t)]' dt
=sec(t)*tan(t)-∫[sec(t)]'*tan(t) dt
=sec(t)*tan(t)-∫[sec(t)*tan(t)]*tan(t) dt
=sec(t)*tan(t)-∫sec(t)*tan^2(t) dt
=sec(t)*tan(t)-∫sec(t)*[sec^2(t)-1] dt
=sec(t)*tan(t)-∫sec^3(t) dt+∫sec(t) dt
=sec(t)*tan(t)-I+ln|sec(t)+tan(t)|+C，
所以2I=sec(t)*tan(t)+ln|sec(t)+tan(t)|+C，
I=sec(t)*tan(t)/2+ln|sec(t)+tan(t)|/2+C，C为任意常数

然后求∫sec^5(t) dt：记J=∫sec^5(t) dt，则J
=∫sec^3(t)*sec^2(t) dt
=∫sec^3(t)*[tan(t)]' dt
=sec^3(t)*tan(t)-∫[sec^3(t)]'*tan(t) dt
=sec^3(t)*tan(t)-∫3sec^2(t)*[sec(t)*tan(t)]*tan(t) dt
=sec^3(t)*tan(t)-3∫sec^3(t)*tan^2(t) dt
=sec^3(t)*tan(t)-3∫sec^3(t)*[sec^2(t)-1] dt
=sec^3(t)*tan(t)-3∫sec^5(t) dt+3∫sec^3(t) dt
=sec^3(t)*tan(t)-3J+3I，
所以4J=sec^3(t)*tan(t)+3I，
J=sec^3(t)*tan(t)/4+3I/4
=sec^3(t)*tan(t)/4+3sec(t)*tan(t)/8+3ln|sec(t)+tan(t)|/8+C，
C为任意常数
综合一下，就可以得出结果，自己算一下。

∫（2x+1）sin2x dx
解：原式=∫2xsin2x dx+∫sin2x dx
=∫xsin2xd2x+1/2∫sin2xd2x
=-∫xdcos2x+1/2∫dcos2x
=-xcos2x+∫cos2xdx+1/2cos2x
=-xcos2x+1/2sin2x+1/2cos2x+c
你那个答案貌似不对，圣诞快乐~

Answer 2

这都是什么啊
根本就看不懂