Evaluate `int
(lnx)/x dx` :
We ` `` `let `u=lnx` . Then `du=1/xdx` and we have:
`intudu=1/2u^2+C` . Substituting for `u` we get `1/2(lnx)^2+C` .
Thus the integral evaluates as `1/2(ln(x))^2+C`
Evaluate `int
(lnx)/x dx` :
We ` `` `let `u=lnx` . Then `du=1/xdx` and we have:
`intudu=1/2u^2+C` . Substituting for `u` we get `1/2(lnx)^2+C` .
Thus the integral evaluates as `1/2(ln(x))^2+C`
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