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Integrate the function
$\sqrt{1+\frac{x^2}{9}}$
$\sqrt{1+\frac{x^2}{9}}$
Solution:
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Verified Answer
$\int \sqrt{1+\frac{\mathrm{x}^2}{9}} \mathrm{dx}=\frac{1}{3} \int \sqrt{\mathrm{x}^2+3^2}$
$=\frac{1}{6}\left[\mathrm{x} \sqrt{\mathrm{x}^2+9}+9 \log \left|\mathrm{x}+\sqrt{\mathrm{x}^2+9}\right|\right]+\mathrm{c}$
$=\frac{1}{6}\left[\mathrm{x} \sqrt{\mathrm{x}^2+9}+9 \log \left|\mathrm{x}+\sqrt{\mathrm{x}^2+9}\right|\right]+\mathrm{c}$
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