Question

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

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

Answer 1

The correct option is D $x{e}^{x+\frac{1}{x}}+c$

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

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

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

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

If $f\left(x\right)=x{e}^{1/x}$

$f^{\prime }\left(x\right)=e^{1/x}-\frac{1}{x}{e}^{1/x}$

$\therefore$ Integral of the form $\int e^x\left(f\left(x\right)f^{\prime }\left(x\right)\right)dx$

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

$=x{e}^{x+1/x}+c$

Hence, solved.