Can do this in a number of ways, here are two of them. The first is to use a substitution:
#color(red)("Method 1")#
#int_1^e (ln(x))/(2x) dx = 1/2 int_1^e (ln(x))/(x) dx#
Let #u = ln(x) implies du = (dx)/x#
Transforming the limits:
#u = ln(x) implies u: 0 rarr 1#
Integral becomes:
#1/2int_0^1 u du =1/2 [1/2u^2]_0^1 = 1/2*1/2 = 1/4#
This is the simpler way, but you might not always be able to make a substitution. An alternative is integration by parts.
#color(red)("Method 2")#
Use integration by parts:
For functions #u(x), v(x)#:
#int uv' dx = uv - int u'v dx#
#u(x) = ln(x) implies u'(x) = 1/x#
#v'(x) = 1/(2x) implies v(x) = 1/2ln(x)#
#int (ln(x))/(2x) dx=1/2ln(x)ln(x) - int (ln(x))/(2x) dx#
Grouping like terms:
#2 int (ln(x))/(2x) dx = 1/2ln(x)ln(x) + C#
#therefore int (ln(x))/(2x) dx = 1/4ln(x)ln(x) + C#
We are working with a definite integral though, so applying limits and removing the constant:
#int_(1)^(e) (ln(x))/(2x) dx = [1/4ln(x)ln(x)]_1^e #
#= 1/4ln(e)ln(e) - 1/4ln(1)ln(1)#
#ln(e) = 1, ln(1) = 0#
#implies int_(1)^(e) (ln(x))/(2x) dx = 1/4#
