Question

Evaluate $\int (1+x-x^{-1})e^{{x+x}^{-1}}dx$

• A. $(x+1)e^{{x+x}^{-1}}+c$
• B. $(x-1)e^{{x+x}^{-1}}+c$
• C. $\frac{1}{2}x{e}^{{x+x}^{-1}}+c$
• D. $x.e^{{x+x}^{-1}}+c$

Answer 1

The correct option is D $x.e^{{x+x}^{-1}}+c$

$\int \left(\right(1+x-\left(\frac{1}{x}\right)).e^{(x+\frac{1}{x})})dx$

$=\int (1+x(1-(\frac{1}{x^2}\left)\right)e^{(x+\frac{1}{x})}dx$

$=\int (e^{(x+(\frac{1}{x})}+x.e^{(x+(\frac{1}{x})}).(1-(\frac{1}{x2}\left)\right)dx$

$=\int \left(\frac{d}{dx}\right(x.e^{(x+(\frac{1}{x})}\left)\right)dx$

$=\int \left(d\right(x.e^{(x+(\frac{1}{x}\left)\right)}\left)\right)$

$=\int \left(d\right(x.e^{(x+(\frac{1}{x}\left)\right)}\left)\right)=x.e^{(x+x^{(-1)})}$

Hence, correct answer is

$x.e^{(x+x^{(-1)})}$+ c