Question

不定积分∫e^(x^2)dx,上限2,下限lnx

1.求不定积分∫(lnx/√1＋x)dx
2.lim (∫e^(t^2) dt)^(-x^2),（这是个定积分,积分上限是X,下限是
x-∞
0,不好打）．

Answer 1

1)令1+x = t²
则dx = 2tdt
∫lnx/√(1+x) dx
=∫ln(t²-1)*2tdt/t
=2∫[ln(t+1) + ln(t-1)]dt
=2∫ln(t+1)d(t+1) + 2∫ln(t-1)d(t-1)
=2[(t+1)ln(t+1) - (t+1) + (t-1)ln(t-1) - (t-1)] + C
=2(t+1)ln(t+1) + 2(t-1)ln(t-1) - 4t + C
2)对极限后面的式子求对数,记为lnA
lnA = -x²ln∫e^t² dt
当x→∞时lnA→0
则A→1
原式=1