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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