積分高手看一下~!!

1. ∫1/√(x²+1)dx= ln| x+√(x²+1) | + C (三角代換可得)
2. ∫√(x²+1) dx =0.5[ x√(x²+1)+ ln| x+√(x²+1) | ]+ C
因 ∫√(x²+1) dx (by I.B.P.)
= x√(x²+1) -∫x²/√(x²+1) dx
= x√(x²+1) -∫√(x²+1) dx +∫1/√(x²+1) dx
故2∫√(x²+1)dx= x√(x²+1) + ln|x+√(x²+1)| + C
3.∫x²√(x²+1) dx (by. I.B.P.)
=(1/3)x³√(x²+1)-(1/3)∫x^4/√(x²+1) dx
=(1/3)x³√(x²+1)-(1/3)∫(x^4+x²)/√(x²+1)dx+(1/3)∫x²/√(x²+1)dx
=(1/3)[x³√(x²+1)-∫x²√(x²+1)dx+∫√(x²+1)dx-∫1/√(x²+1)dx]
移項得(4/3)∫x²√(x²+1)dx
=(1/3)[x³√(x²+1)+∫√(x²+1)dx-∫1/√(x²+1)dx]
故∫x²√(x²+1) dx
=(1/8)[2x³√(x²+1)+x√(x²+1)- ln| x+√(x²+1)| ] + C

三角代換做的出來