Ignoring the 1/2:
Let u = ln x, dv = dx, so that du = (1/x)dx and v = x. Then:
∫ ln x dx = ∫ u dv = uv - ∫ v du
= x ln x - ∫ x (1/x) dx'
∫ ln x dx = x ln x - x + C
Now, just multiply both sides by 1/2 for your integral.
let u=lnx and du=1/2dx then you should get = 1/2[xlnx-x]
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Verified answer
Ignoring the 1/2:
Let u = ln x, dv = dx, so that du = (1/x)dx and v = x. Then:
∫ ln x dx = ∫ u dv = uv - ∫ v du
= x ln x - ∫ x (1/x) dx'
∫ ln x dx = x ln x - x + C
Now, just multiply both sides by 1/2 for your integral.
let u=lnx and du=1/2dx then you should get = 1/2[xlnx-x]