Question

Can someone walk me through the indefinite integral of: t(ln(t+1))?

Answer 1

Rewritten:
\[\int\limits_{ }^{} t \ln(t+1) dt\]

Answer 2

Looks like you could use integration by parts

Answer 3

Yes that's what we are supposed to be exercising, but I don't understand how to do this. What would I use for u, and what would I use for dv?

Answer 4

\[\int t\ln(t+1)dt\]\[u=\ln(t+1)\implies du=\frac{dt}{t+1}\]\[dv=tdt\implies v=\frac12t^2\]

Answer 5

nah maybe not, let me try on paper...

Answer 6

oh yes, it will work the way I did it :)

Answer 7

presumably you got to\[\int t\ln(t+1)dt=\frac t{t+1}-\frac12\int\frac{t^2}{t+1}dt\]and got stuck?

Answer 8

Haha I actually wasn't sure what to pick for u or uv, and why.

Answer 9

dv* not uv

Answer 10

well you have two terms, \(t\) and \(\ln(t+1)\) so one must be u and one must be dv...

Answer 11

you should think to yourself "I don't know how to integrate \(\ln(t+1)\), so it doesn't make a lot of sense to use it for dv since I will have to integrate that"

Answer 12

that means it must be u, which means t must be dv
reasonable?

Answer 13

Okay and ln(t+1) is very easy to find the derivative of. So I should pick it as u

Answer 14

you actually could integrate it, and I bet this would work just fine that way though ;)
bvut I think it is easier to reason it my way
yes u=ln(t+1)

Answer 15

but*

Answer 16

ok so it would look like this:

Answer 17

(t^2 / 2)(ln(t+1) - integral(t/(t+1)) ?

Answer 18

sorry if that looks messy

Answer 19

oh we both messed up :/

Answer 20

\[\int t\ln(t+1)dt=\frac12t^2\ln(t+1)-\frac12\int\frac{t^2}{t+1}dt\]

Answer 21

Where did the 1/2 come from infront of the integral

Answer 22

\[u=\ln(t+1)\implies du=\frac{dt}{t+1}\]\[dv=tdt\implies v=\frac12t^2\]\[\int udv=uv-\int vdu\]\[\int t\ln(t+1)dt=\frac12t^2\ln(t+1)-\frac12\int\frac{t^2}{t+1}dt\]

Answer 23

it came from the fact that v=t^2/2