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April 16, 2014

April 16, 2014

Posted by **Elaine** on Wednesday, February 15, 2012 at 12:15am.

- Calculus -
**Anonymous**, Wednesday, February 15, 2012 at 11:06amlook for a u that does not get too messy on differentiation, and a v that can be integrated without too much troubls

here, it seems logical to use u = arctan(1/x) since we know we're going to get rid of the nasty arctan, and the chain rule will toss in a 1/x^2:

u = arctan(1/x)

du = 1/(1/x^2 + 1)* (-1/x^2) dx

= -1/(1+x^2) dx

dv = dx

v = x

uv-Int v*du = x*arctan(1/x) + Int(x/(1+x^2)) dx

Int v*du = 1/2 * ln(1+x^2)

So, the final answer is x*arctan(1/x) + 1/2 ln(1+x^2)

go ahead -- take the derivative and watch things cancel out!!

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