How do you integrate $int arctan(1/x)$ using integration by parts?

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Answer 1 · 2016-10-17T11:27:37

See the explanation section below.

$int arctan(1/x) dx$

Let $theta = arctan(1/x)$ .

This makes $tan theta = 1/x$ , so $cot theta = x$ .

Furthermore, $dx = -csc^2 theta " " d theta$

The integral becomes:

$int theta (-csc^2 theta) d theta$

Let $u = theta$ and $dv = (-csc^2 theta) d theta$

So $du = d theta$ and $v = cot theta$

$uv-int v du = theta cot theta - int cot theta d theta$

The integral can be found by substitution. We get
$theta cot theta -ln abs sin theta +C$

Using $cot theta = x$ and some trigonometry, we sind $sin theta = 1/sqrt(x^2+1)$

Therefore

$int arctan(1/x) dx = x arctan (1/x)-ln(1/sqrt(x^2+1))+C$

$= x arctan(1/x)+1/2 ln(x^2+1) +C$

How do you integrate int arctan(1/x)int arctan(1/x) using integration by parts?