- #1

karush

Gold Member

MHB

- 3,269

- 5

$a\quad {-\dfrac{1}{3}{(1-\ln{t})}^3+C} \\$

$b\quad {\ln{t}-2\ln{t^2} +\ln{t^3} +C} \\$

$c\quad {-2(1-\ln{t})+C} \\$

$d\quad {\ln{t}-\ln{t^2}+\dfrac{(\ln{t^3})}{3}+C} \\$

$e\quad {-\dfrac{(1-\ln{t^3})}{3}+C}$

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)