变限积分求极限问题,竞赛题,积分区间是[0,n]被积函数是 (x^2)/(x + n^3)求当n趋于正无穷时,积分值是多

∫(x^2)/(x + n^3)dx
=∫[(x+ n^3)^2 -2n^3•x –n^6]/(x + n^3)dx
=∫[(x+ n^3)^2 -2n^3•(x+n^3) +n^6]/(x + n^3)dx
=∫(x+ n^3)dx -2n^3∫dx + n^6∫1/(x + n^3) dx
=∫(x+ n^3)d(x+ n^3) -2n^3∫dx + n^6∫1/(x + n^3) d(x+ n^3)
=(1/2) (x+ n^3)^2 -2n^3•x + n^6•ln(x + n^3)
←→x从0到n
=[(1/2) (n+ n^3)^2 -2n^4 + n^6•ln(n + n^3)] – [(1/2) n^6 +3 n^6•ln n]
=(1/2) •n•(n+2n^3) + n^6•ln[(n + n^3)/n^3]
=(1/2) n^2-n^4 + n^6•ln(1/n^2 + 1)
则
lim(n→∞) (1/2) n^2-n^4 + n^6•ln(1/n^2 + 1)
= ln lim(n→∞) e^[(1/2) n^2-n^4 + n^6•ln(1/n^2 + 1)]
= ln lim(n→∞) (1/n^2 + 1)^( n^6) •e^ [(1/2) n^2-n^4]
= ln lim(n→∞) [(1/n^2 + 1)^(n^2)]( n^4) •e^ [(1/2) n^2-n^4]
= ln lim(n→∞) e^( n^4) •e^ [(1/2) n^2-n^4]
= ln lim(n→∞) e^ [(1/2) n^2-n^4 +n^4]
= ln lim(n→∞) e^ [(1/2) n^2]
= +∞