1. ## arctan integration problem

S arctan(1/x)dx
ok, so i just took my calc 2 exam and there was a definite integral problem on the test and i'm just curious as to how to do it correctly... i typed this in wolfram but i'm not really sure how they achieved the answer. can someone please show me the correct steps to completing this? well, these are the steps that i did.
I decided that integrating by parts would be the most logical choice on my exam..
u=arctan(1/x)
du=1/(x^(-2)+1)
dv=dx
v=x
sooo... xarctan(1/x) - S x/(x^(-2)+1)
thennn i basically did something, which i forgot and got some wackadoo answer that wasn't equal to what my calculator found for the definite integral limits given... HELP!!

2. Originally Posted by sgares
S arctan(1/x)dx

ok, so i just took my calc 2 exam and there was a definite integral problem on the test and i'm just curious as to how to do it correctly... i typed this in wolfram but i'm not really sure how they achieved the answer. can someone please show me the correct steps to completing this? well, these are the steps that i did.
I decided that integrating by parts would be the most logical choice on my exam..
u=arctan(1/x)
du=1/(x^(-2)+1)
dv=dx
v=x
sooo... xarctan(1/x) - S x/(x^(-2)+1)
thennn i basically did something, which i forgot and got some wackadoo answer that wasn't equal to what my calculator found for the definite integral limits given... HELP!!
Letting

$u=\arctan\bigg(\frac{1}{x}\bigg)=arcot(x)$

and $dv=dx$

So $du=\frac{-1}{1+x^2}dx$

and $v=x$

So we have

$\int\arctan\bigg(\frac{1}{x}\bigg)dx=x\cdot{}arcot (x)+\int\frac{x}{x^2+1}dx$

The second integral can be gotten easily

$\int\arctan\bigg(\frac{1}{x}\bigg)dx=x\cdot{}arcot (x)+\frac{1}{2}\ln|x^2+1|+C$

EDIT: And as Krizalid would chastise

since $x^2+1>0\forall{x}\in\mathbb{R}$

We can rewrite this as

$x\cdot{}arcot(x)+\frac{1}{2}\ln(x^2+1)+C$

3. woah, i was off for sure =[ thanks for the reply though!