\int\left\{\frac{1}{\log x}-\frac{1}{(\log x)^{2}}\right\} d x

1.	\int\left\{\frac{1}{\log x}-\frac{1}{(\log x)^{2}}\right\} d x
Answer» ∫ [(1 /log x) - (1 /log²x)] dx = let: log x = t x = e^t dx = e^t dt then, substituting: ∫ [(1 /log x) - (1 /log²x)] dx = ∫ [(1 /t) - (1 /t²)] e^t dt = ∫ [(t - 1) /t²] e^t dt = rewrite this as: ∫ [(t - 1) e^t] t^(- 2) dt = integrate by parts, letting: (t - 1) e^t = u → [(1) e^t + e^t (t - 1)] dt = (e^t + t e^t - e^t) dt = t e^t dt = du t^(- 2) dt = dv → [1/(- 2+1)] t^(- 2+1) = [1/(- 1)] t^(- 1) = - 1 /t = v yielding: ∫ u dv = v u - ∫ v du ∫ [(t - 1) e^t] t^(- 2) dt = (- 1 /t) (t - 1) e^t - ∫ (- 1 /t) t e^t dt = - [(t - 1) /t] e^t + ∫ e^t dt = [(1 - t) /t] e^t + e^t + C = (factoring out e^t) {[(1 - t) /t] + 1} e^t + C = [(1 - t + t) /t] e^t + C = (1 /t) e^t + C = [(e^t) /t] + C let's substitute back log x for t: [e^(log x)] /log x] + C = (being e^log x, a composition of inverses functions, the same as x) (x /log x) + C the answer is: ∫ [(1 /log x) - (1 /log²x)] dx = (x /log x) + C

Answer»

∫ [(1 /log x) - (1 /log²x)] dx =

let:

log x = t

x = e^t

dx = e^t dt

then, substituting:

∫ [(1 /log x) - (1 /log²x)] dx = ∫ [(1 /t) - (1 /t²)] e^t dt =

∫ [(t - 1) /t²] e^t dt =

rewrite this as:

∫ [(t - 1) e^t] t^(- 2) dt =

integrate by parts, letting:

(t - 1) e^t = u → [(1) e^t + e^t (t - 1)] dt = (e^t + t e^t - e^t) dt = t e^t dt = du

t^(- 2) dt = dv → [1/(- 2+1)] t^(- 2+1) = [1/(- 1)] t^(- 1) = - 1 /t = v

yielding:

∫ u dv = v u - ∫ v du

∫ [(t - 1) e^t] t^(- 2) dt = (- 1 /t) (t - 1) e^t - ∫ (- 1 /t) t e^t dt =

- [(t - 1) /t] e^t + ∫ e^t dt =

[(1 - t) /t] e^t + e^t + C =

(factoring out e^t)

{[(1 - t) /t] + 1} e^t + C =

[(1 - t + t) /t] e^t + C =

(1 /t) e^t + C =

[(e^t) /t] + C

let's substitute back log x for t:

[e^(log x)] /log x] + C =

(being e^log x, a composition of inverses functions, the same as x)

(x /log x) + C

the answer is:

∫ [(1 /log x) - (1 /log²x)] dx = (x /log x) + C

Discussion