Question

The integral $\int (1+x-x^{-1})e^{x+x^{-1}}dx$ equals to

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

Answer 1

The correct option is D

$x{e}^{x+x^{-1}}+C$

Explanation for correct option

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

$$\begin{gathered} \Rightarrow \int \left(1+x-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx \\ \Rightarrow \int e^{x+\frac{1}{x}}dx+\int \left(x-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx \\ \Rightarrow x{e}^{x+\frac{1}{x}}-\int \left(1-\frac{1}{x^2}\right)e^{x+\frac{1}{x}}dx+\int \left(x-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx \\ \Rightarrow x{e}^{x+\frac{1}{x}}-\int \left(x-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx+\int \left(x-\frac{1}{x}\right)e^{x+\frac{1}{x}}dx \\ \Rightarrow x{e}^{x+\frac{1}{x}}+C \\ \Rightarrow x{e}^{x+x^{-1}}+C \end{gathered}$$

Hence, option D is correct.