Question

$\int e^x{\left(\frac{x+2}{x+4}\right)}^{2}dx=f\left(x\right)+$ arbitrary constant, then $f(x)$ =

• A. $\frac{x{e}^x}{x+4}$
• B. $\frac{e^x}{x+4}$
• C. $\frac{x{e}^x}{(x+4{)}^2}$
• D. $\frac{e^x}{(x+4{)}^2}$

Answer 1

Answer: (A)

Given, $\int e^x{\left(\frac{x+2}{x+4}\right)}^{2}dx=f(x)+c$
Now, $\int e^x{\left(\frac{x+2}{x+4}\right)}^{2}dx=\int e^x\left(\frac{x^2+4+4x}{(x+4{)}^2}\right)dx$
$=\int e^x\left(\frac{x}{x+4}+\frac{4}{(x+4{)}^2}\right)dx$
Let $g(x)=\frac{x}{(x+4)}$, then $g^{\prime }(x)=\frac{4}{(x+4{)}^2}$
$=\int e^x\left\{g(x)+g^{\prime }(x)\right\}dx$
$=e^{x}g(x)+c=e^x\left(\frac{x}{x+4}\right)+c$
$\therefore f(x)=\frac{x{e}^x}{x+4}$