Blog: `int_1^(sqrt(3)) arctan(1/x) dx` Evaluate the integral

$\displaystyle{\int_{{1}}^{\sqrt{{{3}}}}}{\arctan{{\left(\frac{{1}}{{x}}\right)}}}{\left.{d}{x}\right.}$

If f(x) and g(x) are differentiable functions, then

$\displaystyle\int{f{{\left({x}\right)}}}{g{'}}{\left({x}\right)}{\left.{d}{x}\right.}={f{{\left({x}\right)}}}{g{{\left({x}\right)}}}-\int{f{'}}{\left({x}\right)}{g{{\left({x}\right)}}}{\left.{d}{x}\right.}$

If we write f(x)=u and g'(x)=v, then

$\displaystyle\int{u}{v}{\left.{d}{x}\right.}={u}\int{v}{\left.{d}{x}\right.}-\int{\left({u}'\int{v}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}$

Using the above method of integration by parts,

$\displaystyle\int{\arctan{{\left(\frac{{1}}{{x}}\right)}}}{\left.{d}{x}\right.}={\arctan{{\left(\frac{{1}}{{x}}\right)}}}\cdot\int{1}{\left.{d}{x}\right.}-\int{\left(\frac{{d}}{{\left.{d}{x}}}{\left({\arctan{{\left(\frac{{1}}{{x}}\right)}}}\int{1}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}\right.}$

$\displaystyle={\arctan{{\left(\frac{{1}}{{x}}\right)}}}\cdot{x}-\int{\left(\frac{{1}}{{{1}+{{\left(\frac{{1}}{{x}}\right)}}^{{2}}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left(\frac{{1}}{{x}}\right)}\int{1}{\left.{d}{x}\right.}\right)}{\left.{d}{x}\right.}$

$\displaystyle={x}{\arctan{{\left(\frac{{1}}{{x}}\right)}}}-\int{\left(\frac{{{x}}^{{2}}}{{{{x}}^{{2}}+{1}}}\cdot{\left(-{1}{{x}}^{{-{{2}}}}\right)}\cdot{x}\right)}{\left.{d}{x}\right.}$

$\displaystyle={x}{\arctan{{\left(\frac{{1}}{{x}}\right)}}}+\int\frac{{x}}{{{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}$

Now let's evaluate $\displaystyle\int\frac{{x}}{{{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}$

substitute $\displaystyle{t}={{x}}^{{2}}+{1},\Rightarrow{\left.{d}{t}\right.}={2}{x}{\left.{d}{x}\right.}$

$\displaystyle\int\frac{{x}}{{{{x}}^{{2}}+{1}}}{\left.{d}{x}\right.}=\int\frac{{\left.{d}{t}}}{{{2}{t}}}$

$\displaystyle=\frac{{1}}{{2}}{\ln}{\left|{t}\right|}$

substitute back $\displaystyle{t}={{x}}^{{2}}+{1}$

$\displaystyle=\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{1}\right|}$

$\displaystyle\int{\arctan{{\left(\frac{{1}}{{x}}\right)}}}{\left.{d}{x}\right.}={x}{\arctan{{\left(\frac{{1}}{{x}}\right)}}}+\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{1}\right|}+{C}$

C is a constant

Now evaluate the definite integral,

$\displaystyle{\int_{{1}}^{\sqrt{{{3}}}}}{\arctan{{\left(\frac{{1}}{{x}}\right)}}}{\left.{d}{x}\right.}={{\left[{x}{\arctan{{\left(\frac{{1}}{{x}}\right)}}}+\frac{{1}}{{2}}{\ln}{\left|{{x}}^{{2}}+{1}\right|}\right]}_{{1}}^{\sqrt{{{3}}}}}$

$\displaystyle={\left[\sqrt{{{3}}}{\arctan{{\left(\frac{{1}}{\sqrt{{{3}}}}\right)}}}+\frac{{1}}{{2}}{\ln{{\left({3}+{1}\right)}}}\right]}-{\left[{1}{\arctan{{\left(\frac{{1}}{{1}}\right)}}}+\frac{{1}}{{2}}{\ln{{\left({1}+{1}\right)}}}\right]}$

$\displaystyle={\left[\sqrt{{{3}}}\frac{\pi}{{6}}+\frac{{1}}{{2}}{\ln{{\left({4}\right)}}}\right]}-{\left[\frac{\pi}{{4}}+\frac{{1}}{{2}}{\ln{{2}}}\right]}$

$\displaystyle={\left[\sqrt{{{3}}}\frac{\pi}{{6}}+\frac{{1}}{{2}}{\ln{{\left({{2}}^{{2}}\right)}}}\right]}-{\left[\frac{\pi}{{4}}+\frac{{1}}{{2}}{\ln{{\left({2}\right)}}}\right]}$

$\displaystyle={\left(\sqrt{{{3}}}\frac{\pi}{{6}}+{\ln{{\left({2}\right)}}}-\frac{\pi}{{4}}-\frac{{1}}{{2}}{\ln{{\left({2}\right)}}}\right)}$

$\displaystyle={\left(\sqrt{{{3}}}\frac{\pi}{{6}}-\frac{\pi}{{4}}+\frac{{1}}{{2}}{\ln{{\left({2}\right)}}}\right)}$

$\displaystyle={\left({2}\sqrt{{{3}}}-{3}\right)}\frac{\pi}{{12}}+\frac{{1}}{{2}}{\ln{{\left({2}\right)}}}$