**Q: ∫ x arctanx dx**

A: Integration by parts again: ∫udv = uv – ∫vdu

[Review some older examples of “integration by parts” if you are new to this!]

u = arctanx v = x²/2

du = 1/(1+ x²) dx dv = x dx

So,

= x²/2 (arctanx) – ∫ x²/2 * 1/(1+ x²) dx

Now, let me clean house with the integral part:

∫ x²/2 * 1/(1+ x²) dx = 1/2 ∫ x²/(1+ x²) dx

Now, how to integrate? Hmmm. I am going to do some simplifying with x²/(1+ x²) using polynomial long division:

……… 1

x²+1 | x²

……..-(x² + 1)

…………….-1

Wow, that formating looks whacked, but I will go with it… So:

x²/(1+ x²) = 1 – 1/(1+ x²)

So, back to the problem:

∫ x²/2 * 1/(1+ x²) dx = 1/2 ∫ x²/(1+ x²) dx

= 1/2 ∫ 1 – 1/(1+ x²) dx = 1/2 [∫ 1dx – ∫1/(1+ x²) dx]

= 1/2[x – arctanx]

Now, back to the beginning:

∫ x arctanx dx = x²/2 (arctanx) – ∫ x²/2 * 1/(1+ x²) dx

= x²/2 (arctanx) – 1/2[x – arctanx] + C

thanks for this one, it was a pain in the but for sure

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