Question

please help me find the integral of xlnx dx. i really would appreciate steps not just what rules to use. i know i am using integration by parts. i just get caught in a loop for some reason.

Answer 1

So if you integrate once with u' = x and v = ln x, what do you get?

Answer 2

i did u = lnx

Answer 3

Use my suggestion

Answer 4

ok

Answer 5

so = x and dv = lnx is that how u say to start

Answer 6

u = x*

Answer 7

I'm saying u' = x and v = ln x

Answer 8

if you are originally given xlnx, shouldnt derivative of u be 1

Answer 9

The rule of integration by parts is
\[ \int u' \ v \ dx \ = \ u \ v \ - \ \int u \ v' \ dx \]

Answer 10

yes

Answer 11

We want to 'get rid' of the ln x because it's tricky. Hence it's much better here to start with
u' = x and v=ln x

Answer 12

ok

Answer 13

do you know how to do
\[\int\limits_{}^{}ue^u du\]

Answer 14

probably not

Answer 15

Oh i was going to say to could have put it in this form with a substitution
nvm

Answer 16

this problem in in the intragtion by parts section

Answer 17

ok. with the suggested u' and v, what have you got?

Answer 18

You can use substitution +integration parts if you wanted

Answer 19

@myin: it really is easy to do by parts. I don't think it's helpful to substitute here.

Answer 20

I agree with you James. I was just trying to see if that form was easier for him to integrate by parts.

Answer 21

1/x - integral of

Answer 22

is that the right start?

Answer 23

If u' = x, what is u? And with v=ln x, what is v'?
u = ....
v' = ...

Answer 24

u= 1/2x^2

Answer 25

v = 1/x

Answer 26

u = x^2/2, yes
v = 1/x, yes.

hence the integral equals
\[ \frac{x^2 \ln x}{2} - \int \frac{x^2}{2x} \ dx \]

Answer 27

Now, the new integral simplifies and you can integrate it directly.

Answer 28

v' = 1/x

Answer 29

thanks for your help, im just terrible at math, i just cant grasp this stuff

Answer 30

i get lost when ur splitting the v and v 1's

Answer 31

which is which

Answer 32

u' = x, u = x^2/2
v = ln x, v' = 1/x

Hence
\[ \int u'v = uv - \int uv' \]
\[ \int x \ln x \ dx \ = \ \frac{1}{2}x^2 \ln x - \int \frac{x^2}{2} \frac{1}{x} \ dx \]
yes?

Answer 33

ok

Answer 34

So what's your final answer?

Answer 35

james when ur done here can you jump in my post plz?