int {{{left( {frac{{x + 2}}{{x + 4}}} right)} | vedclass

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(A) We know that $\int {{e^x}[f(x) + f'(x)]dx = {e^x}f(x) + c}$.Let $I = \int {{e^x}{{\left( {\frac{{x + 2}}{{x + 4}}} \right)}^2}dx} $.Rewrite the te

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${e^x}\left( {\frac{x}{{x + 4}}} \right) + c$

Vedclass · Answer

(A) We know that $\int {{e^x}[f(x) + f'(x)]dx = {e^x}f(x) + c}$.Let $I = \int {{e^x}{{\left( {\frac{{x + 2}}{{x + 4}}} \right)}^2}dx} $.Rewrite the term inside the square: $\frac{{x + 2}}{{x + 4}} = \frac{{x + 4 - 2}}{{x + 4}} = 1 - \frac{2}{{x + 4}}$.So,$I = \int {{e^x}{{\left( {1 - \frac{2}{{x + 4}}} \right)}^2}dx} = \int {{e^x}\left( {1 - \frac{4}{{x + 4}} + \frac{4}{{{{(x + 4)}^2}}}} \right)dx} $.Let $f(x) = 1 - \frac{4}{{x + 4}}$. Then $f'(x) = - \frac{d}{{dx}}\left( {\frac{4}{{x + 4}}} \right) = - 4 \cdot \left( { - \frac{1}{{{{(x + 4)}^2}}}} \right) = \frac{4}{{{{(x + 4)}^2}}}$.Thus,$I = \int {{e^x}[f(x) + f'(x)]dx = {e^x}f(x) + c} $.$I = {e^x}\left( {1 - \frac{4}{{x + 4}}} \right) + c = {e^x}\left( {\frac{{x + 4 - 4}}{{x + 4}}} \right) + c = {e^x}\left( {\frac{x}{{x + 4}}} \right) + c$.

$\int {{{\left( {\frac{{x + 2}}{{x + 4}}} \right)}^2}{e^x}\,dx} $ is equal to

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