Integral Calculus
86220
\(\int \frac{(x+3) e^{x}}{(x+4)^{2}} d x\) is equal to
1 \(\frac{e^{x}}{(x+4)}+C\)
2 \(\frac{\mathrm{e}^{\mathrm{x}}}{(\mathrm{x}+4)^{2}}+\mathrm{C}\)
3 \(\frac{\mathrm{e}^{\mathrm{x}}}{(\mathrm{x}+3)}+\mathrm{C}\)
4 \(\frac{1}{(x+4)^{2}}+C\)
Explanation:
(A) : \(I=\int \frac{(x+3) e^{x}}{(x+4)^{2}} d x\)
\(=\int \frac{(\mathrm{x}+4-1) \mathrm{e}^{\mathrm{x}}}{(\mathrm{x}+4)^{2}} \mathrm{dx}=\int \mathrm{e}^{\mathrm{x}}\left(\frac{1}{\mathrm{x}+4}+\frac{-1}{(\mathrm{x}+4)^{2}}\right) \mathrm{dx}\)
\(=e^{x}\left(\frac{1}{x+4}\right)+C\left[\because \int e^{x}\left(f(x)+f^{\prime}(x) d x=e^{x} . f(x)+C\right]\right.\)