Roger Smith

2021-12-30

I have to calculate this integrals
${I}_{1}={\int }_{0}^{2}\frac{\mathrm{arctan}\left(x\right)}{{x}^{2}+2x+2}dx$
${I}_{2}=\underset{n\to \mathrm{\infty }}{lim}{\int }_{0}^{n}\frac{\mathrm{arctan}x}{{x}^{2}+x+1}dx$
I have a hint on the first one, the substitution $x=\frac{2-t}{1+2t}$, but I can't understand how it was conceived.

### Answer & Explanation

Bukvald5z

${I}_{2}={\int }_{0}^{+\mathrm{\infty }}\frac{\mathrm{arctan}x}{{x}^{2}+x+1}dx={\int }_{0}^{1}\frac{\mathrm{arctan}x}{{x}^{2}+x+1}dx+{\int }_{1}^{+\mathrm{\infty }}\frac{\mathrm{arctan}x}{{x}^{2}+x+1}dx$
In the second integral perform the change of variable $y=\frac{1}{x}$
${I}_{2}={\int }_{0}^{1}\frac{\mathrm{arctan}x}{{x}^{2}+x+1}dx+{\int }_{0}^{1}\frac{\mathrm{arctan}\left(\frac{1}{x}\right)}{{x}^{2}+x+1}dx$
$={\int }_{0}^{1}\frac{\left(\mathrm{arctan}x+\mathrm{arctan}\left(\frac{1}{x}\right)\right)}{{x}^{2}+x+1}dx$
$={\int }_{0}^{1}\frac{\frac{\pi }{2}}{{x}^{2}+x+1}dx$
$=\frac{\pi }{2}{\int }_{0}^{1}\frac{1}{{\left(x+\frac{1}{2}\right)}^{2}+\frac{3}{4}}dx$
$=\frac{\pi }{2}×\frac{2}{\sqrt{3}}{\left[\mathrm{arctan}\left(\frac{2x+1}{\sqrt{3}}\right)\right]}_{0}^{1}$
$=\frac{\pi }{\sqrt{3}}\left(\mathrm{arctan}\left(\sqrt{3}\right)-\mathrm{arctan}\left(\frac{1}{\sqrt{3}}\right)\right)$
$=\frac{\pi }{\sqrt{3}}\left(\frac{\pi }{3}-\frac{\pi }{6}\right)\right)$
$=\frac{{\pi }^{2}}{6\sqrt{3}}$
Other important relation for arctan function:
If $x>0$ then,
$\mathrm{arctan}x+\mathrm{arctan}\left(\frac{1}{x}\right)=\frac{\pi }{2}$
If $x<0$ replace $\frac{\pi }{2}$ by $-\frac{\pi }{2}$
And remember too that, for all x real,

Esther Phillips

In ${I}_{1}$ apply the change of variable $x=\frac{2-t}{1+2t}$
Observe that $t=\frac{2-x}{1+2x}$
${I}_{1}={\int }_{0}^{2}\frac{\mathrm{arctan}\left(\frac{2-t}{1+2t}\right)}{{t}^{2}+2t+2}dt$
If $t\in \left[0;2\right]$ then $0<2t<1$ therefore,
${I}_{1}={\int }_{0}^{2}\frac{\mathrm{arctan}\left(2\right)}{{t}^{2}+2t+2}dt-{I}_{1}$
Therefore,
${I}_{1}=\frac{1}{2}{\int }_{0}^{2}\frac{\mathrm{arctan}\left(2\right)}{{t}^{2}+2t+2}dt$
$=\frac{\mathrm{arctan}\left(2\right)}{2}{\left[\mathrm{arctan}\left(x+1\right)\right]}_{0}^{2}$
$=\frac{\mathrm{arctan}\left(2\right)\mathrm{arctan}\left(3\right)}{2}-\frac{\mathrm{arctan}\left(2\right)}{8}\pi$

karton

$\approx 0.2567$

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