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. $-x{e}^{x+x^{-1}}+c$
• D. $x{e}^{x+x^{-1}}+c$

Answer 1

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

We know that;

$\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f\left(x\right)+C$

But there's a more generalized form which deals with functions of $x$ in the exponent;

$\int e^{g\left(x\right)}\left(f\right(x\left)g^{\prime }\right(x)+f^{\prime }(x\left)\right)dx=e^{g\left(x\right)}f\left(x\right)+C$

The above integral has been derived by differentiating $e^{g\left(x\right)}f\left(x\right)$, similar to the derivation of the formula, $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f\left(x\right)+C$.

Hence,

$I=\int (1+x-\frac{1}{x})e^{x+\frac{1}{x}}dxI=\int [x(1-\frac{1}{x^2})+1]e^{x+\frac{1}{x}}dx$

Over here;

$f\left(x\right)=xg\left(x\right)=x+\frac{1}{x}{f}^{\prime }\left(x\right)=1g^{{}^{\prime }}\left(x\right)=1-\frac{1}{x^2}$

Therefore the integral is in the form of;

$\int e^{g\left(x\right)}\left(f\right(x\left)g^{\prime }\right(x)+f^{\prime }(x\left)\right)dx=e^{g\left(x\right)}f\left(x\right)+C$

hence the answer is;

$I=x{e}^{(x+\frac{1}{x})}+C$