I'm studying for a final, and I haven't seen any mention of any problem of this form in class or in my homework. I can't figure out how to go about solving this problem:
$$\int^{e^6}_{1}{\frac{dx}{x(1+\ln(x))}}$$
What I was thinking is:
$$\int{\frac{dx}{x(1+\ln(x))}}+C = \int{\frac{1}{x(1+\ln(x))}dx}+C =ln(ln(x)+1)+C $$
then solve for $$F(e^6) - F(1)$$
But I'm not so sure this is the correct approach. Can someone highlight why the dx is in such an unusual position?
$\endgroup$ 21 Answer
$\begingroup$The $\,dx\,$ term in the numerator is simply replacing the multiple $1$.
$$\int^{e^6}_{1}{\frac{dx}{x(1+\ln(x))}} = \int^{e^6}_{1}{\frac{1}{x(1+\ln(x))}\,dx}$$
This is "sort of like" when we represent a fraction in one of two ways: $\;\dfrac 35 = \dfrac 15\cdot 3.\;$
Of course, we don't mean to imply that "$\,dx\,$" is a number, per se. But you will often see "$dx$" positioned in the numerator, instead of alongside and to the right of the function to be integrated.
ADDED: The result of your integration: $\;F(x) = \ln(\ln(x) + 1) + C,\;$ is "spot on": now you need to simply evaluate $\;F(e^6) - F(1),\;$ which I trust you can do!
$\endgroup$ 3
