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 $$\int \frac{x^4}{1+x^2}\,dx\text{ ?}$$

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Do you mean the integral of $x^3 \tan^{-1} x$? –  George V. Williams Jan 4 '13 at 21:31
Exactly! That's what I'm asking :) –  Syed Sahl Jan 4 '13 at 21:35

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?

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"Addition by zero" is also a helpful trick here: $$\frac{x^4}{1 + x^2} = \frac{x^4 + x^2 - x^2}{1 + x^2} = \frac{x^4 + x^2}{1 + x^2} - \frac{x^2}{1 + x^2} = \frac{x^2(1 + x^2)}{1 + x^2} = x^2 - \frac{x^2}{1 + x^2}$$ The same trick gives: $$\frac{x^2}{1 + x^2} = \frac{x^2 + 1 - 1}{1 + x^2} = \frac{1+x^2}{1+x^2} - \frac{1}{1 + x^2} = 1 - \frac{1}{1+x^2}$$ Hence: $$\frac{x^4}{1 + x^2} = x^2 - \left( 1 - \frac{1}{1+x^2}\right)$$ –  JavaMan Jan 4 '13 at 21:41
@JavaMan: Yes...indeed! –  amWhy Jan 4 '13 at 21:44
@JavaMan Even easier: $x^4-1+1=(x^2-1)(x^2+1)+1$. –  Artem Jan 5 '13 at 2:59
@Artem: Your way is shorter, but I try to invoke as little creativity as possible (so as to maximize the chance that the OP remembers the trick). –  JavaMan Jan 5 '13 at 3:37
$$\frac{x^4}{1+x^2} = x^2 + \frac{1}{1+x^2} -1$$