Question

Evaluate the integral.

Answer 1

\[\int\limits_{}^{}\frac{ \tan^{-1} x}{ x^2 }dx\]

Answer 2

I'm not sure how to set up the beginning for this.

Answer 3

my guess is parts

Answer 4

so if I choose \[ u = \frac{ 1 }{ x^2 }, du = -\frac{ 1 }{ x }dx\]
dv = arctanx, what is v?

Answer 5

just cause the anti derivative of \(\frac{1}{x^2}\) is easy
as is the derivative of \(\tan^{-1}(x)\)
it is just a guess

Answer 6

do it the other way

Answer 7

actually you have a mistake there

Answer 8

\[u = arctanx, du = \frac{ 1 }{ 1+x^2 }, dv = \frac{ 1 }{ x^2 }, v = -\frac{ 1 }{ x }\]

Answer 9

yeah that would be my guess

Answer 10

Okay, I will try it.

Answer 11

at least you get a rational function to deal with, arctangent goes. so that should work

Answer 12

I think you can do a partial fraction decomposition on \[\int\limits vdu\]