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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**Further Reading**