What is the definite integral of arctan(1/x) evaluated from 1 to sqrt(3)?

Question

Does u =arctan and du =1/1+x^2 and
dv=1/x and v =ln|x|? if so why do I keep getting the wrong answer?

Answer 1

No, that's completely incorrect. Arctan is not being multiplied by 1/x, rather 1/x is being used as the input to the arctangent function. It's the difference between f(x)*g(x) and f(g(x)), with this case being the latter. As such, there is no way to separate the 1/x from the arctangent -- they MUST be considered as a single unit.

Doing this problem correctly, we let u=arctan (1/x), du = 1/(1+(1/x)²) * (-1/x²) dx = -1/(x²+1) dx (using the chain rule), dv = dx, v = x. Then we have:

[1, √3]∫arctan (1/x) dx
x arctan (1/x) |[1, √3] + [1, √3]∫x/(x²+1) dx
√3 arctan (1/√3) - 1 arctan 1 + [1, √3]∫x/(x²+1) dx
π√3/6 - π/4 + [1, √3]∫x/(x²+1) dx

Now making the substitution u=x²+1, du=2x dx, x dx = du/2, x=1 ⇒ u=2, x=√3 ⇒ u=4, we have:

π√3/6 - π/4 + 1/2 [2, 4]∫1/u du
π√3/6 - π/4 + 1/2 ln |u| |[2, 4]
π√3/6 - π/4 + 1/2 (ln 4 - ln 2)
π√3/6 - π/4 + 1/2 ln (4/2)
π√3/6 - π/4 + ln (√2)

And we are done.