MHB 4.1.1 AP calculus Exam Int with U substitution

[karush]

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)

 

[skeeter]

substitution ...

and (a) is correct, verified by its derivarive

 

[HOI]

Since you titled this "int with u substitution" it looks like you have chosen substitution!

Yes, let u= 1- ln(t). Then du= -dt/t so the integral becomes
$\int u^2(-du)= -u^3/3+ C= -1/3(1- ln(t))^3+ C.

That is (a) as you say. Well done!

 

[karush]

wow I was expecting about 6 steps (think you left off a \$ sign)

$\displaystyle\int u^2(-du)= -u^3/3+ C= -1/3(1- ln(t))^3+ C$