1.求不定积分∫(lnx/√1＋x)dx
1个回答

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

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