I have to use integration by parts to integrate tan^-1 (1/x)dx. I'm not sure what to use as u or dv. We were taught to use uv-integral of (vdu) to do this.
Calculus - Anonymous, Wednesday, February 15, 2012 at 11:06am
look 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!!