Question

$\int (1+x-x^{-1})e^{x+x^{-1}}dx$ is equal to

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

Answer 1

The correct option is C $x{e}^{x+x^{-1}}+C$

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

$=\int [x.e^{x+x^{-1}}(1-\frac{1}{x^2})+e^{x+x^{-1}}]dx$ $\left[\because \int x{f}^{\prime }\left(x\right)+f\left(x\right)dx=xf\left(x\right)+C\right]$

$\therefore \int (1+x-x^{-1})e^{x+x^{-1}}dx=x{e}^{x+x^{-1}}+C$