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Seeking indefinite integral ∫ (LNX) DX
∫(lnx)dx
=xlnx-∫xdlnx
=xlnx-∫dx
=xlnx-x+C
Partial integral (e ^ x + LNX) DX
The original formula is e ^ xdx + lnxdx = e ^ xdx + (xlnxd (LNX)) = e ^ x + xlnx-1,
∫e^(-x+lnx)dx
So we can find this indefinite integral,
∫e^(-x+lnx)dx
=∫xe^(-x)dx
=-∫xd(e^(-x))
=-(xe^(-x)-∫e^(-x)dx)
=-(xe^(-x)-(-e^(-x)))+c
=-(x+1)(e^(-x))+c
∫（lnx/x^2）*（e^lnx）dx=
The original formula = ∫ LNX / xdx = ∫ lnxd (LNX)
1/2(lnx)²+c
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