Question

FInding integrals by using the integration of parts.
integral of xln(x)dx=

Answer 1

I have x^2 ln (x)- integral(x^2)dln(x))= x^2 ln(x)- integral(x^3/3 times by 1/x)

Answer 2

oo a fun integral! :D

Answer 3

more like brutal ;~;

Answer 4

It looks like this is what you were trying to do with your parts, but something got messed up in there.\[\Large u=\ln x \qquad\to\qquad du=\frac{1}{x}dx\]\[\Large dv=x\;dx \qquad\to\qquad v=\frac{1}{2}x^2\]

Answer 5

\[\Large \bf \int\limits u dv \quad=\quad uv-\int\limits v\;du\]

Answer 6

Understand where the 1/2 is coming from?

Answer 7

i thought when you find the integral of x^2 dont you add one to the exponent and then put it underneath?

Answer 8

We don't have an x^2 to integrate though.
We have an x^1 to integrate.

Answer 9

isnt it x^2 at the beginning?

Answer 10

Hmm? at the beginning of what?
At the beginning of our solution?
No there would be a 1/2 connected to it, right?\[\Large \bf\color{#CC0033}{dv=x\;dx}\qquad \to\qquad \color{#CC0033}{v=\frac{1}{2}x^2}\]

Answer 11

when i first evaluated it dont you take the x from dx and mutiply it by the xln(x)? itd give you x^2ln(x)?

Answer 12

Hmm I'm not quite sure what you mean :(

Answer 13

Maybe the colors will help D: Maybe not.. we'll see..
\[\Large \bf\int\limits \color{orangered}{\ln x}(\color{royalblue}{x\;dx})\]\[\Large\bf =\quad \color{royalblue}{\frac{1}{2}x^2}\color{orangered}{\ln x}-\int\limits \color{royalblue}{\frac{1}{2}x^2}\color{orangered}{\frac{1}{x}dx}\]

Answer 14

\[\Large \bf\int\limits\color{orangered}{u}(\color{royalblue}{dv})\]\[\Large\bf\color{royalblue}{v}\color{orangered}{u}-\int\limits\color{royalblue}{v}\color{orangered}{du}\]

Answer 15

too confusing? :c

Answer 16

D: where are you getting 1/2?

Answer 17

how are you familiar with integration by parts?
Let's try to set it up in a way that makes sense to you.

Are you used to using `u` and `dv` or something else?
Do you usually ignore the differential dx? would that make things easier? :o

Answer 18

Like if I said u=ln x, and dv=x
is that easier to understand? :o

Answer 19

ummm. no lol sorry.

Answer 20

Anyway, to answer your question.
We assigned our `x` to be `dv`.
To find `v` we have to integrate.
\[\Large \int\limits x^1\;dx \quad=\quad \frac{1}{1+1}x^{1+1} \quad=\quad \frac{1}{2}x^2\]