Single Variable Calculus, Chapter 8, 8.1, Section 8.1, Problem 26

Evaluate $\displaystyle \int^{\sqrt{3}}_1 \arctan \left( \frac{1}{x} \right) dx$ by using Integration by parts.
If we let $\displaystyle u = \arctan \left( \frac{1}{x} \right)$ and $dv = dx$, then
$\displaystyle du = \frac{\frac{d}{dx}\left(\frac{1}{x}\right)}{1 + \left( \frac{1}{x} \right)^2} = \frac{-\frac{1}{x^2}}{1 + \frac{1}{x^2}} = \frac{-1}{x^2+1} dx \text{ and } v = x$

So,

$
\begin{equation}
\begin{aligned}
\text{so, } \int^{\sqrt{3}}_1 \arctan \left( \frac{1}{x} \right) dx = uv - \int vdu &= x \arctan \left( \frac{1}{x} \right) - \int x \left( \frac{-1}{x^2+1} \right) dx\\
\\
&= x \arctan \left( \frac{1}{x} \right) + \int \frac{x}{x^2+1} dx

\end{aligned}
\end{equation}
$


To evaluate $\displaystyle \int \frac{x}{x^2+1} dx$, we let $u_1 = x^2 +1$ so $du_1 = 2x dx$

$
\begin{equation}
\begin{aligned}
\int \frac{x}{x^2 + 1} dx = \int \frac{1}{u_1} \left( \frac{du_1}{2} \right) = \frac{1}{2} \int \frac{du_1}{u_1} &= \frac{1}{2} \ln u_1\\
\\
&= \frac{1}{2} \ln \left( x^2 + 1 \right)
\end{aligned}
\end{equation}
$


Going back to the first equation,
$\displaystyle \int^{\sqrt{3}}_1 \arctan \left( \frac{1}{x} \right) dx = x \arctan \left( \frac{1}{x} \right) + \frac{1}{2} \ln (x^2 +1)$
Evaluating from $x = 1$ to $x = \sqrt{3}$;
$\displaystyle \int^{\sqrt{3}}_1 \arctan \left( \frac{1}{x} \right) = \frac{\sqrt{3}\pi}{6} - \frac{\pi}{4} + \ln \sqrt{2}$