Dillard

2021-10-20

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

### Answer & Explanation

Alannej

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,
${\int }_{\sqrt{3}}^{3}={\int }_{\sqrt{3}}^{3}\frac{dt}{{t}^{2}+{\left(\sqrt{3}\right)}^{2}}$
$=\left[\frac{1}{\sqrt{3}}{\mathrm{tan}}^{-1}{\left(\frac{t}{\sqrt{3}}\right]}_{\sqrt{3}}^{3}$
$=\left[\frac{1}{\sqrt{3}}\frac{\pi }{3}-\frac{1}{\sqrt{3}}{\mathrm{tan}}^{-1}\left(1\right)\right]$
$=\left[\frac{1}{\sqrt{3}}\left(\frac{\pi }{3}-\frac{\pi }{4}\right]$
$=\frac{1}{\sqrt{3}}\frac{\pi }{12}$
$=\frac{\pi }{12\sqrt{3}}$
$=\frac{\sqrt{3}\pi }{36}$

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