第1个回答 2010-04-21
1. ∫1/√x+1 dx
=∫x^(-1/2)+1dx
=2x^(1/2)+x+C
2. 原式=∫[e^(4x)]^(1/2)dx
=(2/3)*(1/4)*[e^(4x)]^(3/2)+C
=(1/6)[e^(4x)]^(3/2)+C
3. 原式=∫x(1-4x^2)^(1/2)dx
=(2/3)*(-8)*(1-4x^2)^(3/2)+C
=-(16/3)(1-4x^2)^(3/2)+C
4. 原式=∫cos[x^(1/2)]dx
设x^(1/2)=a, x=a^2, dx=2ada
原式=∫2acosada
=2asina-∫2sinada
=2asina+2cosa
=2x^(1/2)sin[x^(1/2)]+2cos[x^(1/2)]