Integrate the function in Exercise. 1+ x^2 9 - Vidyadip

Question

Integrate the function in Exercise.
$\sqrt{1+\frac{\text{x}^2}{9}}$

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Answer

$\int\sqrt{1+\frac{\text{x}^2}{9}}\text{dx}=\int\sqrt{\frac{9+\text{x}^2}{9}}\text{dx}=\int\frac{\sqrt{\text{x}^2+3^2}}{3}\text{dx}=\frac{1}{3}\int\sqrt{\text{x}^2+3^2}\text{dx}$
$=\frac{1}{3}\Bigg[\frac{\text{x}}{2}\sqrt{\text{x}^2+3^2}+\frac{3}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+3^2}\Big|\Bigg]+\text{c}\Bigg[\therefore\ \int\sqrt{\text{x}^2+\text{a}^2}\text{dx}=\frac{\text{x}}{2}\sqrt{\text{x}^2+\text{a}^2}+\frac{\text{a}^2}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+\text{a}^2}\Big|\Bigg]+\text{c}$
$=\frac{\text{x}}{6}\sqrt{\text{x}^2+9}+\frac{3}{2}\text{log}\Big|\text{x}+\sqrt{\text{x}^2+9}\Big|+\text{c}$

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