Question

when integrating radt times lnt by parts should u be lnt or rad t?

Answer 1

thats a little tooooo cryptic, can you english it up some?

Answer 2

sqrt(t).ln(t) perhaps?

Answer 3

\(u=\ln t\)

Answer 4

LN(t) looks alot like int

Answer 5

\[\int\limits_{1}^{\sqrt{5}}\sqrt{t}lnt\]dt

Answer 6

Here 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

Answer 7

ok thanks...so lets say dv=lntdt. what is v then? the antiderivative of lnt i mean?

Answer 8

Well \(\sqrt{t}\) is algebraic. So we choose L which is \(u=\ln t\) and \(dv=\sqrt{t}dt\)

Answer 9

tlnt-t is the integral of lnt. To find that, you need to use by parts