int {frac{{(x + 3){e^x}}}{{{{(x + 4)}^2}}},dx | vedclass

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Question

(C) We need to evaluate the integral $I = \int {\frac{{(x + 3){e^x}}}{{{{(x + 4)}^2}}}dx}$.Rewrite the numerator $(x + 3)$ as $(x + 4 - 1)$:$I = \int

Vedclass · Accepted Answer

$\frac{{{e^x}}}{{x + 4}} + c$

Vedclass · Answer

(C) We need to evaluate the integral $I = \int {\frac{{(x + 3){e^x}}}{{{{(x + 4)}^2}}}dx}$.Rewrite the numerator $(x + 3)$ as $(x + 4 - 1)$:$I = \int {{e^x}\frac{{(x + 4) - 1}}{{{{(x + 4)}^2}}}dx}$$I = \int {{e^x}\left( {\frac{{x + 4}}{{{{(x + 4)}^2}}} - \frac{1}{{{{(x + 4)}^2}}}} \right)dx}$$I = \int {{e^x}\left( {\frac{1}{{x + 4}} - \frac{1}{{{{(x + 4)}^2}}}} \right)dx}$Recall the standard integral formula: $\int {{e^x}(f(x) + f'(x))dx = {e^x}f(x) + c}$.Here,let $f(x) = \frac{1}{{x + 4}}$. Then $f'(x) = -\frac{1}{{{{(x + 4)}^2}}}$.Thus,$I = {e^x}\left( {\frac{1}{{x + 4}}} \right) + c = \frac{{{e^x}}}{{x + 4}} + c$.

$\int {\frac{{(x + 3){e^x}}}{{{{(x + 4)}^2}}}\,dx} = \,$

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