int frac{(x+3) e^{x}}{(x+4)^{2}} d x is equ | vedclass

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(C) We are given the integral $ I = \int \frac{(x+3) e^{x}}{(x+4)^{2}} d x $. To solve this,we rewrite the numerator as $(x+4-1)$: $ I = \int e^{x} \f

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$ \frac{e^{x}}{(x+4)}+C $

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(C) We are given the integral $ I = \int \frac{(x+3) e^{x}}{(x+4)^{2}} d x $. To solve this,we rewrite the numerator as $(x+4-1)$: $ I = \int e^{x} \frac{(x+4-1)}{(x+4)^{2}} d x $. Splitting the fraction,we get: $ I = \int e^{x} \left[ \frac{x+4}{(x+4)^{2}} - \frac{1}{(x+4)^{2}} \right] d x $. $ I = \int e^{x} \left[ \frac{1}{x+4} - \frac{1}{(x+4)^{2}} \right] d x $. We know the standard integration formula: $ \int e^{x} (f(x) + f'(x)) d x = e^{x} f(x) + C $. Here,let $ f(x) = \frac{1}{x+4} $. Then,$ f'(x) = -\frac{1}{(x+4)^{2}} $. Substituting these into the formula,we get: $ I = e^{x} \left( \frac{1}{x+4} \right) + C = \frac{e^{x}}{x+4} + C $.

$ \int \frac{(x+3) e^{x}}{(x+4)^{2}} d x $ is equal to

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