The value of $\int \mathrm{e}^x\left(\frac{x^2+4 x+4}{(x+4)^2}\right) \mathrm{d} x$ is

The value of $\int \mathrm{e}^x\left(\frac{x^2+4 x+4}{(x+4)^2}\right) \mathrm{d} x$ is
  1. $\mathrm{e}^x\left(\frac{x}{x+4}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  2. $\mathrm{e}^x\left(\frac{4}{x+4}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  3. $\mathrm{e}^x\left(\frac{x}{(x+4)^2}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  4. $\mathrm{e}^x\left(\frac{4}{(x+4)^2}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.

Solution

$\begin{aligned} & \int \mathrm{e}^x\left[\frac{x^2+4 x+4}{(x+4)^2}\right] \mathrm{d} x \\ & =\int \mathrm{e}^x\left[\frac{x(x+4)+4}{(x+4)^2}\right] \mathrm{d} x \\ & =\int \mathrm{e}^x\left[\frac{x}{x+4}+\frac{4}{(x+4)^2}\right] \mathrm{d} x\end{aligned}$ $\begin{aligned}=\mathrm{e}^x\left(\frac{x}{x+4}\right)+\mathrm{c} & \\ & \cdots\left[\because \int \mathrm{e}^x\left[\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right] \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}\right]\end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 2)

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