Integrate (lnx)/x^2 dx between limits 1 and 5
Let I = integral[(lnx)/x^2 dx] for simplicity.Firstly, we realise we must use integration by parts. This is:Integral [u(x)v'(x) dx] = u(x)v(x) - Integral[u'(x)v(x) dx]So we can see that, by letting u(x)=lnx and v'(x)=1/x^2, we have:I = (lnx)(-1/x) - integral[(1/x)(-1/x) dx] = -(lnx)/x + integral [1/x^2 dx] = -(lnx)/x - 1/x (+C would be used for indefinite integral; where there are no limits)Plug in the limits, we have:-ln5/5-1/5+ln1/1+1/1=4/5 - (ln5)/5or (4-ln5)/5
JL
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