Question

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

(a) $\frac{e^x}{x+4}+C$

(b) $\frac{e^x}{x+3}+C$

(c) $\frac{1}{{\left(x+4\right)}^2}+C$

(d) $\frac{e^x}{{\left(x+4\right)}^2}+C$

Answer 1

(a) $\frac{e^x}{x+4}+C$

$$\begin{gathered} \mathrm{Let}I=\int \frac{\left(x+3\right)}{{\left(x+4\right)}^2}{e}^{x}dx \\ \Rightarrow \int \left[\frac{x+4-1}{{\left(x+4\right)}^2}\right]e^{x}dx \\ \Rightarrow \int \left[\frac{1}{\left(x+4\right)}-\frac{1}{{\left(x+4\right)}^2}\right]e^{x}dx \\ \mathrm{As},\mathrm{we}\mathrm{know}\mathrm{that}\int e^x\left\{f\left(x\right)+f\text{'}\left(x\right)\right\}dx=e^{x}f\left(x\right)+C \\ \therefore I=\frac{e^x}{x+4}+C \end{gathered}$$