<s>[INST] Determine the indefinite integral for $x\tan^{-1}x^{2}$ I'm not sure how to move on after using integration by parts to arrive at 
$$\frac{x^{2}\tan^{-1}x^{2}}{2} - \frac{1}{2}\int \frac{x^{2}}{x^{4} + 1}\,dx$$
 [/INST] $$\int x\arctan x^2\ dx=\frac{1}{2}x^2\arctan x^2 -\int \frac{x^3}{1+x^4}\ dx$$
by integration by parts, where u=arctan(x^2) and dv=xdx.
next use a simple u-substition where $u=1+x^4$.
$$\int \arctan x^2\ dx=\frac{1}{2} x^2 \arctan x^2-\int \frac{x^3}{1+x^4}\ dx
             =\frac{1}{2}x^2\arctan x^2 -\frac{1}{4}\ln(1+x^4)+C  $$
qed
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