- #1
karush
Gold Member
MHB
- 3,269
- 5
Evaluate $\displaystyle\int{\dfrac{{(1-\ln{t})}^2}{t} dt=}$
$a\quad {-\dfrac{1}{3}{(1-\ln{t})}^3+C} \\$
$b\quad {\ln{t}-2\ln{t^2} +\ln{t^3} +C} \\$
$c\quad {-2(1-\ln{t})+C} \\$
$d\quad {\ln{t}-\ln{t^2}+\dfrac{(\ln{t^3})}{3}+C} \\$
$e\quad {-\dfrac{(1-\ln{t^3})}{3}+C}$
ok we can either expand the numerator or go with u substitution $u=1-\ln{t}$
Just by intution I would quess the answer is (a)
$a\quad {-\dfrac{1}{3}{(1-\ln{t})}^3+C} \\$
$b\quad {\ln{t}-2\ln{t^2} +\ln{t^3} +C} \\$
$c\quad {-2(1-\ln{t})+C} \\$
$d\quad {\ln{t}-\ln{t^2}+\dfrac{(\ln{t^3})}{3}+C} \\$
$e\quad {-\dfrac{(1-\ln{t^3})}{3}+C}$
ok we can either expand the numerator or go with u substitution $u=1-\ln{t}$
Just by intution I would quess the answer is (a)