Having one of the factors go to zero is a special case, so usually you apply the tabular method and keep differentiating/integrating until the product of the factors is something you can integrate or deal with, I will elaborate with some examples:
say we'd like to integrate $I=\int \:\ln\left(x\right)\:dx$ using integration by parts and the tabular method:
$$\begin{pmatrix}u&dv\\ \ln\left(x\right)&1\\ \frac{1}{x}&x\end{pmatrix}$$
you stop after one time because you know that you can integrate $\int \:x\cdot \frac{1}{x}\:dx$
and your answer becomes $I=x\:\cdot \ln\left(x\right)-\int \:x\cdot \frac{1}{x}dx$
remember that the sign is alternating between terms.
Now lets do $\int \:e^x\sin\left(x\right)\:dx$
$$\begin{pmatrix}u&dv\\ \sin\left(x\right)&e^x\\ \cos\left(x\right)&e^x\\ -\sin\left(x\right)&e^x\end{pmatrix}$$
we stop after two times and get:
$\int \:e^x\sin\left(x\right)\:dx= e^x\sin\left(x\right)-e^x\cos\left(x\right)+-\int \:e^x\sin\left(x\right)\:dx$
by adding $\int \:e^x\sin\left(x\right)\:dx$ to both sides we get:
$2\int \:e^x\sin\left(x\right)\:dx=e^x\sin\left(x\right)-e^x\cos\left(x\right)+C_0$
$\int \:e^x\sin\left(x\right)\:dx=\frac{e^x}{2}\left(\sin\left(x\right)-\cos\left(x\right)\right) +C$
Best Answer
I think the tabular method is mainly integration by parts done several steps at once, and in this particular case it doesn't really make things different as there's only one step:
$$I:=\int x\arctan x\,dx\;\;\;\;\;,\;\;\;\begin{cases}u=\arctan x&,\;\;u'=\frac1{1+x^2}\\{}\\v'=x&,\;\;v=\frac12x^2\end{cases}\;\;\implies$$
$$I=\frac12x^2\arctan x-\frac12\int\frac{x^2}{1+x^2}dx=\frac12x^2\arctan x-\frac12\int\left(1-\frac1{1+x^2}\right)dx=$$
$$\frac12\left(x^2\arctan x-x+\arctan x\right)=\frac12\left((1+x^2)\arctan x-x\right)$$