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The integral is of the form $\int e^x [f(x) + f'(x)] dx$, which equals $e^x f(x) + C$. We need to manipulate the given expression to fit this form.
Let's rewrite the integral as: $\int e^{x}[\frac{1}{(1+x^{2})^{\frac{3}{2}}}+\frac{x}{\sqrt{1+x^{2}}}]dx = \int e^x \left[ \frac{1}{(1+x^2)^{3/2}} + \frac{x}{(1+x^2)^{1/2}} \right] dx$
Let $f(x) = \frac{x}{\sqrt{1+x^2}}$. Then we need to find $f'(x)$. $f'(x) = \frac{d}{dx} \left( \frac{x}{\sqrt{1+x^2}} \right)$
Using the quotient rule: $f'(x) = \frac{\sqrt{1+x^2} \cdot 1 - x \cdot \frac{1}{2}(1+x^2)^{-1/2} \cdot 2x}{1+x^2} = \frac{\sqrt{1+x^2} - \frac{x^2}{\sqrt{1+x^2}}}{1+x^2} = \frac{1+x^2 - x^2}{(1+x^2)^{3/2}} = \frac{1}{(1+x^2)^{3/2}}$
Now we have $\int e^x [f(x) + f'(x)] dx = \int e^x \left[ \frac{x}{\sqrt{1+x^2}} + \frac{1}{(1+x^2)^{3/2}} \right] dx$. Since $f(x) = \frac{x}{\sqrt{1+x^2}}$ and $f'(x) = \frac{1}{(1+x^2)^{3/2}}$, the integral is $e^x f(x) + C = e^x \frac{x}{\sqrt{1+x^2}} + C$
Final Answer: $e^{x}\frac{x}{\sqrt{1+x^{2}}}+C$
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