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How do you integrate #int 1/(xlnx)# using substitution?

1 Answer

Jul 28, 2016

# ln(ln(x)) + C#

Explanation:

Let #u=ln(x) implies du = 1/xdx#

So

#int (dx)/(xln(x)) = int (du)/(u)#

#int (du)/(u) = ln(u) + C = ln(ln(x)) + C#

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