I'm supposed to take the integral of $\arctan(x)$ but get the wrong answer. I know my answer is wrong because I'm supposed to do integration by parts and consider the integral to be $ \int {\arctan(x) \cdot 1 dx} $. However, my question is WHY do I consider it to be $ \arctan(x) \cdot 1 dx $ and take these parts:
$$u = \arctan(x)$$
$$du = \frac {1} {x^2+1} $$
$$dv = dx$$
$$v = x?$$
For example, if I do $ \int {x^2 dx} $ I don't have to do integration by parts of $\int {x^2 \cdot 1 dx}$?
This question became really messy so here's a TLDR: Why do I consider $\int {\arctan(x) dx}$ to be equal to $\int {\arctan(x) 1 dx}$?
It is simply a trick to apply the integration by parts $\int fg'= fg -\int f'g$ with $g'=1$.
Note that a similar trick is used for
$$\int \log x dx=\int 1\cdot \log x dx= x\log x - \int x\cdot \frac1x dx$$
we assume $g'(x)=1$ as such that $g(x)=x$.