Search any question & find its solution
Question:
Answered & Verified by Expert
The value of $\int \mathrm{e}^x\left(\frac{x^2+4 x+4}{(x+4)^2}\right) \mathrm{d} x$ is
Options:
Solution:
1891 Upvotes
Verified Answer
The correct answer is:
$\mathrm{e}^x\left(\frac{x}{x+4}\right)+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
$\begin{aligned} & \int \mathrm{e}^x\left[\frac{x^2+4 x+4}{(x+4)^2}\right] \mathrm{d} x \\ & =\int \mathrm{e}^x\left[\frac{x(x+4)+4}{(x+4)^2}\right] \mathrm{d} x \\ & =\int \mathrm{e}^x\left[\frac{x}{x+4}+\frac{4}{(x+4)^2}\right] \mathrm{d} x\end{aligned}$
$\begin{aligned}=\mathrm{e}^x\left(\frac{x}{x+4}\right)+\mathrm{c} & \\ & \cdots\left[\because \int \mathrm{e}^x\left[\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right] \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}\right]\end{aligned}$
$\begin{aligned}=\mathrm{e}^x\left(\frac{x}{x+4}\right)+\mathrm{c} & \\ & \cdots\left[\because \int \mathrm{e}^x\left[\mathrm{f}(x)+\mathrm{f}^{\prime}(x)\right] \mathrm{d} x=\mathrm{e}^x \mathrm{f}(x)+\mathrm{c}\right]\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.