Evaluate the following integrals: ^3_01 x^2+9 dx - Vidyadip

Question

Evaluate the following integrals:$\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$

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Answer

$\int\limits^3_0\frac{1}{\text{x}^2+9}\text{ dx}$$=\int\limits^3_0\frac{1}{\text{x}^2+3^2}\text{ dx}$
$=\frac{1}{3}\Big[\tan^{-1}\frac{\text{x}}{3}\Big]^3_0$
$=\frac{1}{3}\big(\tan^{-1}1-\tan^{-1}0\big)$
$=\frac{1}{3}\Big(\frac{\pi}{4}-0\Big)$
$=\frac{\pi}{12}$

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