Integration of $x^3 \tan^{-1}(x)$ by parts I'm having problem with this question. How would one integrate $$\int x^3\tan^{-1}x\,dx\text{ ?}$$
After trying too much I got stuck at this point. How would one integrate $$\int \frac{x^4}{1+x^2}\,dx\text{ ?}$$

You've done the hardest part. Now, the problem isn't so much about "calculus"; you simply need to recall what you've learned in algebra:
$(1)$ Divide the numerator of the integrand: $\,{x^4}\,$ by its denominator, $\,{1+x^2}\,$ using *polynomial long division *, (linked to serve as a reference).
This will give you:
$$\int \frac{x^4}{1+x^2}\,dx = \int (x^2 + \frac{1}{1+x^2} -1)\,dx=\;\;?$$

Alternatively: Notice also that $$\int \frac{x^4}{x^2 + 1}\,dx= \int \frac{[(x^4 - 1) + 1]}{x^2 + 1}\,dx$$
$$= \int \frac{(x^2 + 1)(x^2 - 1) + 1}{x^2 + 1} \,dx = \int x^2 - 1 + \frac{1}{x^2 + 1}\,dx$$
I trust you can take it from here?
