Question

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

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

Answer 1

The correct option is B

$x{e}^{\left(x+\frac{1}{x}\right)}+C$

Explanation for the correct answer:

Finding the value for the given integral:

$$\begin{gathered} \int \left(1+x-\frac{1}{x}\right)\left(e^{x+\frac{1}{x}}\right)dx=\int \left(1+x\left(1-\frac{1}{x^2}\right)\right)\left(e^{x+\frac{1}{x}}\right)dx \\ =\int \left(e^{x+\frac{1}{x}}\right)+x\left(1-\frac{1}{x^2}\right)\left(e^{x+\frac{1}{x}}\right)dx \\ \text{Using integration by parts},u=x,v=\left(1-\frac{1}{x^2}\right)\left(e^{x+\frac{1}{x}}\right) \\ \int udv=uv-\int vdu\left[\text{Integrating}\int \left(e^{x+\frac{1}{x}}\right).\left(1-\frac{1}{x^2}\right)dx \\ =\int \left(e^{x+\frac{1}{x}}\right)d\left(x+\frac{1}{x}\right) \\ =\int \left(e^{x+\frac{1}{x}}\right)\right] \\ =\int \left(e^{x+\frac{1}{x}}\right)+x\left(e^{x+\frac{1}{x}}\right)-\int \left(e^{x+\frac{1}{x}}\right)\frac{d\left(x\right)}{dx} \\ =\int \left(e^{x+\frac{1}{x}}\right)+x\left(e^{x+\frac{1}{x}}\right)-\int \left(e^{x+\frac{1}{x}}\right)dx+C \\ =x\left(e^{x+\frac{1}{x}}\right)+C \end{gathered}$$

Hence, option (B) is the correct answer.