求X^2*根号（x^2+1）的不定积分,麻烦各位了.可以的话尽量用分部积分公式.

Question

求X^2*根号（x^2+1）的不定积分,麻烦各位了.
可以的话尽量用分部积分公式.

Answer 1

x=tant,t=arctanx
dx=(sect)^2 dt
S X^2*根号（x^2+1）dx
=S (tant)^2*sect *(sect)^2 dt
=S[(sect)^2-1]*(sect)^3 dt
=S(sect)^5 *dt-S(sect)^3*dt
首先求∫sec^3(x) dx：记I=∫sec^3(x) dx,则I
=∫sec(x)*sec^2(x) dx
=∫sec(x)*[tan(x)]' dx
=sec(x)*tan(x)-∫[sec(x)]'*tan(x) dx
=sec(x)*tan(x)-∫[sec(x)*tan(x)]*tan(x) dx
=sec(x)*tan(x)-∫sec(x)*tan^2(x) dx
=sec(x)*tan(x)-∫sec(x)*[sec^2(x)-1] dx
=sec(x)*tan(x)-∫sec^3(x) dx+∫sec(x) dx
=sec(x)*tan(x)-I+ln|sec(x)+tan(x)|+C,
所以2I=sec(x)*tan(x)+ln|sec(x)+tan(x)|+C,
I=sec(x)*tan(x)/2+ln|sec(x)+tan(x)|/2+C,C为任意常数
然后求∫sec^5(x) dx：记J=∫sec^5(x) dx,则J
=∫sec^3(x)*sec^2(x) dx
=∫sec^3(x)*[tan(x)]' dx
=sec^3(x)*tan(x)-∫[sec^3(x)]'*tan(x) dx
=sec^3(x)*tan(x)-∫3sec^2(x)*[sec(x)*tan(x)]*tan(x) dx
=sec^3(x)*tan(x)-3∫sec^3(x)*tan^2(x) dx
=sec^3(x)*tan(x)-3∫sec^3(x)*[sec^2(x)-1] dx
=sec^3(x)*tan(x)-3∫sec^5(x) dx+3∫sec^3(x) dx
=sec^3(x)*tan(x)-3J+3I,
所以4J=sec^3(x)*tan(x)+3I,
J=sec^3(x)*tan(x)/4+3I/4
=sec^3(x)*tan(x)/4+3sec(x)*tan(x)/8+3ln|sec(x)+tan(x)|/8+C,
C为任意常数
再把t=arctanx代入即可