Answer
$c+\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{4n+3}}{(4n+3)(2n+1)}$
$R=1$
Work Step by Step
$arctan(x^{2})=\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{4n+2}}{(2n+1)}$
$=x^{2}-\frac{1}{3}x^{6}+\frac{1}{5}x^{10}-\frac{1}{7}x^{14}+......+\frac{(-1)^{n}x^{4n+2}}{2n+1}+...$
Now,
$\int arctan(x^{2})dx=(x^{2}-\frac{1}{3}x^{6}+\frac{1}{5}x^{10}-\frac{1}{7}x^{14}+......+\frac{(-1)^{n}x^{4n+2}}{2n+1}+...)dx$
$=c+\Sigma_{n=0}^{\infty}\dfrac{(-1)^{n}x^{4n+3}}{(4n+3)(2n+1)}$
$R=1$
