Question

integral 0 to 1 ln(x)dx

Answer 1

integrate by parts du=dx, u=lnx

Answer 2

dv=dx, u=lnx
I mean...

Answer 3

so just taking the derivative of ln(x)
but isnt it 1/x?

Answer 4

yep

Answer 5

then after getting the derivtive just put the 0 to 1 limit?

Answer 6

what? I said integrate by parts, do you know how to do that?

Answer 7

Um, \[\int \ln(x)dx\ne\frac{1}{x}+C\] D:

Answer 8

@ivanmlerner I hope you're not typing the full solution

Answer 9

No, relax

Answer 10

but for imporoper integrals, we have to to take the limits

Answer 11

yes, but first you gotta integrate it, the limits come in at the evaluation

Answer 12

im trying to explain the integration by parts from scratch

Answer 13

okay, don't let me stop you. sorry to interrupt but I didn't want you to jump the gun.

Answer 14

im not asking or answers, i am just trying to understand how to set it up :(

Answer 15

Integration by parts is the analogous (not sure if this word exists in english) of the product rule for derivatives.
Product rule: \[\frac{ d(f(x)g(x)) }{ dx }=\frac{ df }{ dx }g+\frac{ dg }{ dx }f\]
Integrating that in x, we get: \[f(x)g(x)=\int\limits_{}^{}\frac{ df }{ dx }g dx+\int\limits_{}^{}\frac{ dg }{ dx }f dx\]
When you put one of the integrals to the other side you get: \[f(x)g(x)-\int\limits\limits_{}^{}\frac{ df }{ dx }g dx=\int\limits\limits_{}^{}\frac{ dg }{ dx }f dx\]
Now the only problem, is to find g and f so that it becomes simpler to solve.

Answer 16

its \[\int\limits_{1}^{0}lnxdx\]

Answer 17

so use what @ivanmlerner showed you and sub in
u=lnx
dv=dx

Answer 18

the way I write integration by pars, which is the same as ivans, is\[\int udv=uv-\int vdu\]

Answer 19

i was asking someone else XD
but thankx. i'll try the integration by parts

Answer 20

sorry, I didn't see you ask anything :/