求X^2*根号（x^2+1）的不定积分，麻烦各位了。。

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

∫x^2√(x^2+1)dx
=∫√(x^2+1)^3dx-∫√(x^2+1)dx

∫√(x^2+1)dx
=x√(x^2+1)-∫x^2dx/√(x^2+1)
=x√(x^2+1)-∫√(x^2+1)dx +∫dx/√(x^2+1)
其中∫dx/√(x^2+1) x=tanu, dx=du/(cosu)^2
cosu=1/√(1+x^2) sinu=x/√(1+x^2)
=∫du/cosu=∫dsinu/[(1-sinu)(1+sinu)]
=(1/2)ln|(1+sinu)/(1-sinu)|
=ln|(1+sinu)/cosu|
=ln|(x+√(x^2+1)|=ln(x+√(1+x^2)
=x√(x^2+1)+ln(x+√(1+x^2) - ∫√(x^2+1)dx
2∫√(x^2+1)dx=x√(x^2+1)+ln(x+√(1+x^2))
∫√(x^2+1)dx=x√(x^2+1)/2+(1/2)ln(x+√(1+x^2))

∫√(1+x^2)^3dx
=x√(1+x^2)^3-∫3x^2√(1+x^2)dx

∫x^2√(1+x^2)dx= x√(1+x^2)^3+x√(x^2+1)/2 +(1/2)ln(x+√(1+x^2))+C0-3∫x^2√(1+x^2)dx
4∫x^2√(1+x^2)dx=x√(1+x^2)^3+x√(1+x^2)/2+(1/2)ln(x+√(1+x^2)+C0
∫x^2√（1+x^2)dx=(1/4)x√(1+x^2)^3+(1/8)x√(1+x^2)+(1/8)ln(x+√(1+x^2))+C