Question
What is the value of $$\int_1^2 {{e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)} dx\,?$$
A.
$$e\left( {\frac{e}{2} - 1} \right)$$
B.
$$e\left( {e - 1} \right)$$
C.
$$e - \frac{1}{e}$$
D.
$$0$$
Answer :
$$e\left( {\frac{e}{2} - 1} \right)$$
Solution :
$$\eqalign{
& {\text{Let }}I = \int_1^2 {{e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)} dx \cr
& = \int_1^2 {{e^x}\left( {f\left( x \right) + f'\left( x \right)} \right)} dx\,\,\,\,\,\,\,\left[ {{\text{where}}\,f\left( x \right) = \frac{1}{x}} \right] \cr
& = {e^x}\left. {f\left( x \right)} \right|_1^2 \cr
& \therefore \,I = \left. {\frac{{{e^x}}}{x}} \right|_1^2 = \frac{{{e^2}}}{2} - e = e\left( {\frac{e}{2} - 1} \right) \cr} $$