Question

换元积分法求xln²x分之1 e²到e³的定积分？

Answer 1

令 令 u = lnx， 则 x = e^u, dx = e^udu,
I = ∫<e^2, e^3>x(lnx)^2dx = ∫<2, 3>e^u(u^2)e^udu
= ∫<2, 3>u^2e^(2u)du = (1/2)∫<2, 3>u^2de^(2u)
= (1/2){[u^2e^2u]<2, 3> - 2∫<2, 3>ue^(2u)du}
= (9e^6-4e^4)/2 - (1/2)∫<2, 3>ude^(2u)
= (9e^6-4e^4)/2 - (1/2)[ue^2u]<2, 3> + (1/2)∫<2, 3>e^(2u)du
= (9e^6-4e^4)/2 - (3e^6-2e^4)/2 + (1/4)[e^(2u)]<2, 3>
= (9e^6-4e^4)/2 - (3e^6-2e^4)/2 + (e^6-e^4)/4
= (13/4)e^6 - (5/4)e^4

I

Answer 2

=1/6

方法如下，
请作参考：