infinite series integration $x^2 \tan^{-1} x^3$, why can't you directly integrate the infinite series representation? We have $$(\arctan(x^3))' = \frac{3x^2}{1+x^6} = \frac{3x^2}{1-(-x^6)} = 3x^2\sum_{n=0}^{\infty} (-x^6)^n.$$
Integrating this gives $$3\sum_{n=0}^{\infty} \frac{(-1)^n x^{6n+5}}{6n+3}.$$
Apparently it's correct except for the $\frac{1}{6n+3}$ which should be $(2n+1)$, if i remember correctly. But why? With many series you can directly use the integration rules you know, so why not here?
Why should I here write out the original infinite series and then integrate part by part and then make an infinite equation out of that?

Well,
$$\begin{align}
\int 3x^2 \sum_{n=0}^\infty (-x^6)^n \,\textrm{d}x
& = \int 3x^2 \sum_{n=0}^\infty x^{12n} \,\textrm{d}x + \int -3x^8 \sum_{n=0}^\infty x^{12n} \,\textrm{d}x \\ 
& = 3 \int \sum_{n=0}^\infty x^{12n + 2} \,\textrm{d}x - 3\int \sum_{n=0}^\infty x^{12n + 8} \,\textrm{d}x \\ 
& = 3 \sum_{n=0}^\infty \int x^{12n+2} \,\textrm{d}x - 3 \sum_{n=0}^\infty \int x^{12n+8} \,\textrm{d}x \\
& = \sum_{n=0}^\infty \frac{3x^{12n+3}}{12n+3} - \sum_{n=0}^\infty \frac{3x^{12n+9}}{12n+9} = \textrm{arctan}(x^3)\end{align}.
$$
Or even shorter:
$$\begin{align}
\int 3x^2\sum_{n \ge 0} (-x^6)^n \,\textrm{d}x & = 3\sum_{n \ge 0} \int (-x^6)^n\cdot x^2 \,\textrm{d}x = \sum_{n \ge 0} \frac{3 x^3 (-x^6)^n}{6n+3} \\ & = \sum_{n \ge 0}\frac{(-1)^n (x^3)^{2n+1}}{2n+1} = \arctan (x^3).
\end{align}$$
