The value of int e^x left( frac{x^2+4x+4}{(x+ | vedclass

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(A) We need to evaluate the integral $I = \int e^x \left( \frac{x^2+4x+4}{(x+4)^2} \right) dx$. First,rewrite the numerator: $x^2+4x+4 = x(x+4) + 4$.

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$e^x \left( \frac{x}{x+4} \right) + c$,where $c$ is a constant of integration.

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(A) We need to evaluate the integral $I = \int e^x \left( \frac{x^2+4x+4}{(x+4)^2} \right) dx$. First,rewrite the numerator: $x^2+4x+4 = x(x+4) + 4$. So,the integral becomes: $I = \int e^x \left( \frac{x(x+4) + 4}{(x+4)^2} \right) dx$ $I = \int e^x \left( \frac{x}{x+4} + \frac{4}{(x+4)^2} \right) dx$. Let $f(x) = \frac{x}{x+4}$. Then $f'(x) = \frac{(x+4)(1) - x(1)}{(x+4)^2} = \frac{x+4-x}{(x+4)^2} = \frac{4}{(x+4)^2}$. Since the integral is of the form $\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$,we have $I = e^x \left( \frac{x}{x+4} \right) + c$.

The value of $\int e^x \left( \frac{x^2+4x+4}{(x+4)^2} \right) dx$ is

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