integral of ∫1/(x(1+lnx)), limit 1 bottom, e^7 on top. thanx.?

Answer 1

= ln 8

Using this rule:

∫ f'(x)/f(x) dx = ln f(x) + C

You can do it by recognition. the 1/x term is the derivative of (1+ln x).

Thus

∫ dx / [x(1 + ln x)] = ln(1 + ln x) + C

Apply the limits = ln (1 + 7) - ln (1 + 0) = ln 8

Answer 2

Put 1+lnx= z so 1/x dx = dz
and you get the integral between 1 and 8
of dz/z = ln(z) between 1 and 8
= ln(8)