Verified answer

All you have to do is integrate with the variable and set this equal to the value we want. First, antidifferentiate:

(a) [ln|t|] on 1 to x = ln(5)

Then plug in x and 1 following F(b) - F(a) from the fundamental theorem:

= ln(x) - ln(1) = ln(5)

= ln(x) = ln(5)

x = 5

(b) Same integral:

[ln|t|] on 1 to x = 1

= ln(x) - 0 = 1

= ln(x) = 1

x = e

(ln(1) = 0 and ln(e) = 1 are facts about logs you probably should remember, if you don't already)