Question

Evaluate xlnxdx

Answer 1

Please explain

Answer 2

Is there a image to go with this?

Answer 3

\[\huge\rm \int\limits_{ }^{} x \cdot ln(x)~ dx\]

Answer 4

You mean, you could make sense of that?

Answer 5

u-sub will work
u=something whose derivative is already inside the function

Answer 6

yes i think so lol..

Answer 7

let u =lnx
\[\large\rm \color{blue}{u= \ln(x) } ~~~~~,~~~~ du=\frac{1}{x} \color{Red}{ dx}\]
solve du=1/x dx for dx\[\large\rm dx = x~ du\]
hmmm when we plugin x du for dx the other x will not cancel out so hmm
try integration by parts

Answer 8

integration parts: when we integrate two function (and when u -sub) doesn't work we apply the IBP formula \[\huge\rm u \cdot v - \int\limits_{}^{} v~ du \]
`LIATE` is helpful to figure out which function will be the u
L: logs
I:Inverse trig
A: algebraic (x, x^n )
T: Trig
E: Exponential (e^x)
so for example if we have to integrate ( trig times exponential ) functions
u will be equal to trig because it comes first in the above list.

Answer 9

integration by parts*

Answer 10

\[\large\rm \int\limits_{ }^{} x \cdot \color{ReD}{\ln(x) }~~dx\]
algebraic times log (according to `LIATE` log comes first in the list) therfore
`u=lnx`\[\large\rm u = \ln x ~~~\]
whatever left after picking the function for u is going to be the `dv` \[\large\rm dv= x ~~dx\]
integrate dv to find v
and take the derivative of u to find du
and then plug these into the IBP formula

Answer 11

*clap*