The value of the integral \( \mathrm{I}=\int\displaystyle\frac{e^{x}(x-2)}{(x+1)^{4}}\,dx \) is,
[\( C \) is an arbitrary constant]
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{e^{x}}{(x+1)^{3}}+C \)
Since \( x-2=(x+1)-3 \), the fraction splits into two parts:
\( \displaystyle I=\int e^{x}\left[\frac{1}{(x+1)^{3}}-\frac{3}{(x+1)^{4}}\right]dx \)
This is of the form \( \int e^{x}\left[f(x)+f'(x)\right]dx=e^{x}f(x)+C \) with
\( \displaystyle f(x)=\frac{1}{(x+1)^{3}},\qquad f'(x)=-\frac{3}{(x+1)^{4}} \)
\( \displaystyle I=\frac{e^{x}}{(x+1)^{3}}+C \)
Shortcut: Differentiate Option (2):
\( \displaystyle \frac{d}{dx}\left[\frac{e^{x}}{(x+1)^{3}}\right]=\frac{e^{x}(x+1)^{3}-3e^{x}(x+1)^{2}}{(x+1)^{6}} \)
\( \displaystyle =\frac{e^{x}(x-2)}{(x+1)^{4}} \)