a math

a)∫(4+(1/x))^2dx
=∫16+8/x+(1/x))^2dx
=16x+8lnx-(3/x))^3

b)∫x^2e^(x3) dx
=∫1/3d[e^(x3)]
=1/3[e^(x3)]

c)∫(Inx) /x dx
=∫(Inx) d(lnx)
=[(lnx)^2]/2

d)∫√x in(x)dx
=x√x inx-∫[x(1/2)(1/√x)lnx+x√x(1/x)]dx
=x√x inx-∫[(1/2)√xlnx)+√x]dx
=x√x inx-∫(1/2)√xlnx)dx+(2/3)(x)^(3/2)
∫√x in(x)dx=x√x inx-∫(1/2)√xlnx)dx+(2/3)(x)^(3/2)
3/2∫√x in(x)dx=x√x inx+(2/3)(x)^(3/2)
so ∫√x in(x) dx
=2/3x√x inx+(4/9)(x)^(3/2)

2011-05-10 23:11:52 補充：
唔好意思，四題都係寫漏+C
因為係indefinite integration

a) ∫(4+(1/x))^2dx
=∫16+8x^(－1)+x^(－2)dx
=16x+8ln｜x｜－1/x+c

b) ∫x^2e^(x3) dx
=(1/3) ∫e^(x3) dx^3
=(1/3)e^(x^3)+c

c) ∫( lnx) /x dx
= ∫( lnx)dlnx
=(lnx)^2/2+c

d) ∫√x ln(x) dx
Set u=lnx，dv=x^(1/2)dx
du=(1/x)dx，v=(2/3)x^(3/2)
∫√x ln(x) dx
=lnx*(2/3)x^(3/2)－∫(2/3)x^(3/2)(1/x)dx
=(2/3)lnx*x^(3/2)－(2/3)∫x^(1/2)dx
=(2/3)lnx*x^(3/2)－(2/3)*(2/3)x^(3/2)+c
=(2/3)lnx*x^(3/2)－(4/9)x^(3/2)+c