Dillard

2021-10-20

Evaluate the integrals

$\int}_{\sqrt{3}}^{3}\frac{dt}{3+{t}^{2}$

Alannej

Skilled2021-10-21Added 104 answers

We have to evaluate the integral

$\int}_{\sqrt{3}}^{3}\frac{dt}{3+{t}^{2}$

We know

$\int \frac{dx}{{x}^{2}+{a}^{2}}=\frac{1}{a}{\mathrm{tan}}^{-1}\left(\frac{x}{a}\right)$

Therefore,

$\int \frac{dx}{{x}^{2}+{a}^{2}}=\frac{1}{a}{\mathrm{tan}}^{-1}\left(\frac{x}{a}\right)$

Therefore,

$\int}_{\sqrt{3}}^{3}={\int}_{\sqrt{3}}^{3}\frac{dt}{{t}^{2}+{\left(\sqrt{3}\right)}^{2}$

$=[\frac{1}{\sqrt{3}}{\mathrm{tan}}^{-1}{\left(\frac{t}{\sqrt{3}}\right]}_{\sqrt{3}}^{3}$

$=[\frac{1}{\sqrt{3}}\frac{\pi}{3}-\frac{1}{\sqrt{3}}{\mathrm{tan}}^{-1}\left(1\right)]$

$=[\frac{1}{\sqrt{3}}(\frac{\pi}{3}-\frac{\pi}{4}]$

$=\frac{1}{\sqrt{3}}\frac{\pi}{12}$

$=\frac{\pi}{12\sqrt{3}}$

$=\frac{\sqrt{3}\pi}{36}$

We know

Therefore,

Therefore,

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