\( \int\displaystyle\frac{(x-3)e^x}{(x-1)^3}\,dx \) is equal to:
(where C is an arbitrary constant)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{e^x}{(x-1)^2}+C \)
The integrand can be brought to the form \( e^x\{f(x)+f'(x)\} \), whose integral is \( e^xf(x)+C \). Writing x − 3 = (x − 1) − 2:
\( \displaystyle \frac{(x-3)e^x}{(x-1)^3}=e^x\left\{\frac{x-1}{(x-1)^3}-\frac{2}{(x-1)^3}\right\} \)
\( \displaystyle =e^x\left\{\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}\right\} \)
With \( f(x)=\frac{1}{(x-1)^2} \):
\( \displaystyle f'(x)=-\frac{2}{(x-1)^3} \)
So the integrand is \( e^x\{f(x)+f'(x)\} \), and
\( \displaystyle \int\frac{(x-3)e^x}{(x-1)^3}\,dx=\frac{e^x}{(x-1)^2}+C \)
Shortcut: Differentiate Option (4) by the quotient rule:
\( \displaystyle \frac{d}{dx}\left(\frac{e^x}{(x-1)^2}\right)=\frac{e^x(x-1)^2-2(x-1)e^x}{(x-1)^4} \)
\( \displaystyle =\frac{e^x(x-1-2)}{(x-1)^3} \)
\( \displaystyle =\frac{(x-3)e^x}{(x-1)^3} \)