Integral [(log_10_x)/x] dx?

Question

(Log base 10) ...please explain

Answer 1

Note that log_10 x = ln x / ln 10, and ln 10 is a constant.

So ∫ (log_10 x) / x dx
= (1/ln 10) ∫ (ln x / x) dx, let u = ln x, du = dx.1/x
= (1/ln 10) ∫ u du
= (1/ln 10) (u^2 / 2) + const
= (ln x)^2 / (2 ln 10) + const.

Answer 2

Let y = log x (Log to the base 10
then
x = 10^y = e^z
y ln(10) = z = ln(x)
log(x)ln(10) = ln(x)
log(x) = ln(x)/ln(10)

∫[(log(x))/x]dx =
(1/ln(10)∫[(ln(x))/x]dx =
now, let u = ln(x), du = dx/x
(1/ln(10)∫udu =
(1/(2ln(10))u^2 + C =
(1/(2ln(10))(ln(x))^2 + C =
(1/(2ln(10))(log(x)ln(10))^2 + C =

(ln(10)/2)(log(x))^2 + C