Question

Use integration by parts to find the following indefinite integral

Answer 1

\[\int\limits_{}^{}x^2 \ln x dx\] where u = lnx and dv = x^2

Answer 2

so v would = \[\frac{ 1 }{ 3 }x^3 dx?\]

Answer 3

and du = 1/x

Answer 4

Yup, now just plug it into the formula, do you know what it is?

Answer 5

yeah so it will look like this:\[\ln(x) (\frac{ 1 }{ 3 }x^3 dx) - \int\limits_{}^{} (lnx) (x^2 dx)\]

Answer 6

or is that only somewhat right

Answer 7

do I need to do it again?

Answer 8

isn't the formula uv - integral(u dv)?

Answer 9

It's: \(\Large \int u dv = uv-\int v du\)

Answer 10

ah yes, I messed up my handwriting with the u and the v

Answer 11

so wouldn't the last part be 1/x?

Answer 12

Yea, the equation would turn out to be \(\Large \frac{lnx}{3} x^3 - \frac{1}{3} \int x^2 dx \) if I'm not mistaken.

Answer 13

The x^3 and 1/x will reduce to x^2, and we can just pull that 1/3 out

Answer 14

okay, let me write this out

Answer 15

Just integrate and you'll be set :)

Answer 16

I am just trying to figure out how you got there

Answer 17

okay I gotcha

Answer 18

You sure? I can try and explain it more if you need?

Answer 19

\[\frac{ 1 }{ 3 } lnx - \frac{ 1 }{ 9 } x^3 + c\]

Answer 20

You forgot your x^3 on the left side, but other than that perfect ~