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If $\int e^x\left(\frac{x+2}{x+4}\right)^2 d x=f(x)+$ arbitrary constant, then $f(x)=$
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Verified Answer
The correct answer is:
$\frac{x e^x}{x+4}$
Given, $\int e^x\left(\frac{x+2}{x+4}\right)^2 d x=f(x)+\mathcal{c}$
Now,
$$
\begin{aligned}
& \int e^x\left(\frac{x+2}{x+4}\right)^2 d x=\int e^x\left(\frac{x^2+4+4 x}{(x+4)^2}\right) d x \\
& =\int e^x\left(\frac{x}{x+4}+\frac{4}{(x+4)^2}\right) d x
\end{aligned}
$$
$$
\text { Let } \begin{aligned}
g(x)= & \frac{x}{(x+4)^{\prime}} \text {, then } g^{\prime}(x)=\frac{4}{(x+4)^2} \\
& =\int e^x\left\{g(x)+g^{\prime}(x)\right\} d x \\
& =e^x g(x)+c=e^x\left(\frac{x}{x+4}\right)+c
\end{aligned}
$$
$$
\therefore \quad f(x)=\frac{x e^x}{x+4}
$$
Now,
$$
\begin{aligned}
& \int e^x\left(\frac{x+2}{x+4}\right)^2 d x=\int e^x\left(\frac{x^2+4+4 x}{(x+4)^2}\right) d x \\
& =\int e^x\left(\frac{x}{x+4}+\frac{4}{(x+4)^2}\right) d x
\end{aligned}
$$
$$
\text { Let } \begin{aligned}
g(x)= & \frac{x}{(x+4)^{\prime}} \text {, then } g^{\prime}(x)=\frac{4}{(x+4)^2} \\
& =\int e^x\left\{g(x)+g^{\prime}(x)\right\} d x \\
& =e^x g(x)+c=e^x\left(\frac{x}{x+4}\right)+c
\end{aligned}
$$
$$
\therefore \quad f(x)=\frac{x e^x}{x+4}
$$
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