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anonymous
 5 years ago
when integrating radt times lnt by parts should u be lnt or rad t?
anonymous
 5 years ago
when integrating radt times lnt by parts should u be lnt or rad t?

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0thats a little tooooo cryptic, can you english it up some?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sqrt(t).ln(t) perhaps?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0LN(t) looks alot like int

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{1}^{\sqrt{5}}\sqrt{t}lnt\]dt

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Here is a trick that most of the time worsk. Use I L A T E for choosing u. I(inverse), L(og), A(lgebraic), T(rig), E(xp). In your problem you have A(lgebraic) and L(og). Since L come first before A. Choose L as your u. To learn more look at here: http://www.youtube.com/watch?v=cdLF_gymRI

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok thanks...so lets say dv=lntdt. what is v then? the antiderivative of lnt i mean?

watchmath
 5 years ago
Best ResponseYou've already chosen the best response.0Well \(\sqrt{t}\) is algebraic. So we choose L which is \(u=\ln t\) and \(dv=\sqrt{t}dt\)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0tlntt is the integral of lnt. To find that, you need to use by parts
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