why not try both and see what you learn
there is no "correct" vs. "incorrect", both will work. It's just a matter of which turns out to be easier. I suspect you are doing it correctly you just got some algebra or something wrong. If you want post precisely what you have done and I'll see if I can figure out where you went wrong.
Assuming that you meant $\displaystyle \begin{align*} dv = t\,dt \end{align*}$, I agree with your setting of the functions, so you should get
$\displaystyle \begin{align*} \int{ t\ln{(t + 1)} \, dt} &= \frac{t^2}{2}\ln{(t + 1 )} - \int{\frac{t^2\,dt}{t + 1}} \\ &= \frac{t^2}{2}\ln{(t + 1)} - \int{ \frac{ \left( u - 1 \right) ^2 \, du }{u} } \textrm{ after letting } u = t + 1 \implies du = dt \\ &= \frac{t^2}{2} \ln{(t + 1)} - \int{ \frac{ \left( u^2 - 2u + 1 \right) \, du}{u} } \\ &= \frac{t^2}{2}\ln{(t + 1)} - \int{ u - 2 + \frac{1}{u} \, du } \\ &= \frac{t^2}{2}\ln{(t + 1)} - \left( \frac{u^2}{2} - 2u + \ln{|u|} \right) + C \\ &= \frac{t^2}{2} \ln{(t + 1)} - \frac{ \left( t + 1 \right) ^2}{2} + 2 \left( t + 1 \right) - \ln{ \left| t + 1 \right| } + C \end{align*}$
[QUOTE=romsek;818121]there is no "correct" vs. "incorrect", both will work. It's just a matter of which turns out to be easier. I suspect you are doing it correctly you just got some algebra or something wrong. If you want post precisely what you have done and I'll see if I can figure out where you went wron
I will try a similar problem on Monday and post my complete reply. Enjoy your weekend.
[QUOTE=nycmath;818127]for my edification I wrote up a quick sheet to verify that both methods work
The first line computes the integral so we know what the answer is.
then we integrate by parts with u=t, etc.
At the end of the first bit we take the difference of the integral and the integral found by integrating by parts and see that it is 0.
In the second part we flip u and dv and show the same thing.