Determine the indefinite integral for $x\tan^{-1}x^{2}$

Question

I'm not sure how to move on after using integration by parts to arrive at

$$\frac{x^{2}\tan^{-1}x^{2}}{2} - \frac{1}{2}\int \frac{x^{2}}{x^{4} + 1}\,dx$$

Answer 1

$$\int x\arctan x^2\ dx=\frac{1}{2}x^2\arctan x^2 -\int \frac{x^3}{1+x^4}\ dx$$

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

Answer 2

Since $$\int \arctan(x)\,dx = x\arctan x-\int \frac{x}{1+x^2}\,dx = x \arctan(x)-\frac{1}{2}\log(1+x^2) $$ we clearly have: $$\int x\arctan(x^2)\,dx = \frac{x^2}{2}\arctan(x^2)-\frac{1}{4}\log(1+x^4).$$

Answer 3

Do not integrate by parts immediately. Firstly take x^2 = k The x will vanish on substitution Then integrate by parts.When integrating differentiate tan^-1 (x) and integrate 1

source: engineering student

Answer 4

Don't integrate by parts right away. Note that $d(x^2)=2x\,dx$. Let $u = x^2$

$$ \int x\arctan x^2 dx = \frac{1}{2} \int\arctan u\,du$$

You can take it from here.