Question

The value of integral $\int{\left( 1+x-\dfrac{1}{x} \right){{e}^{x+\dfrac{1}{x}}}dx}$ is equal to

& A.\left( x-1 \right){{e}^{x+\dfrac{1}{x}}}+c \\\ & B.x{{e}^{x+\dfrac{1}{x}}}+c \\\ & C.\left( x+1 \right){{e}^{x+\dfrac{1}{x}}}+c \\\ & D.-x{{e}^{x+\dfrac{1}{x}}}+c \\\ \end{aligned}$$

Answer 1

To solve this question, we will use the formula of integral which is given as
\int{{{e}^{x}}}\left\\{ f\left( x \right)+f'\left( x \right) \right\\}dx={{e}^{x}}f\left( x \right)+c
Where, f'(x) is the derivative of f(x).
We will use the function $f(x)\text{ as f}\left( x \right)=x{{e}^{\dfrac{1}{x}}}$ and use the formula above to get answer. Also, we will use the product rule of differentiation as:
$\dfrac{d}{dx}\left( h\left( x \right)-g\left( x \right) \right)=h\left( x \right)\dfrac{d}{dx}g\left( x \right)+\dfrac{d}{dx}\left( h\left( x \right) \right)-g\left( x \right)$

Complete step-by-step answer:
Given that, $\int{\left( 1+x-\dfrac{1}{x} \right){{e}^{x+\dfrac{1}{x}}}dx}$
Let $I=\int{\left( 1+x-\dfrac{1}{x} \right){{e}^{x+\dfrac{1}{x}}}dx}$
Now, write ${{e}^{x+\dfrac{1}{x}}}={{e}^{x}}{{e}^{\dfrac{1}{x}}}$
Using this in above value of I, we get:
$I=\int{\left( 1+x-\dfrac{1}{x} \right){{e}^{x}}{{e}^{\dfrac{1}{x}}}dx}$
Now, taking ${{e}^{x}}$ common one and multiplying ${{e}^{\dfrac{1}{x}}}$ inside in all terms we get:
$I=\int{\left( {{e}^{x}}\left( {{e}^{\dfrac{1}{x}}}+x{{e}^{\dfrac{1}{x}}}-\dfrac{1}{x}{{e}^{\dfrac{1}{x}}} \right) \right)dx}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}$
Now, assume $f\left( x \right)=x{{e}^{\dfrac{1}{x}}}$
We will differentiate f(x) in respect to x. To do so, we will apply product rule of differentiation which says that,
$\dfrac{d}{dx}\left( h\left( x \right).g\left( x \right) \right)=h\left( x \right)\dfrac{d}{dx}g\left( x \right)+\dfrac{d}{dx}\left( h\left( x \right) \right).g\left( x \right)$
Using product rule of differentiation taking
$h\left( x \right)=x\text{ and g}\left( x \right)={{e}^{\dfrac{1}{x}}}$ we get

& \dfrac{d}{dx}f\left( x \right)=f'\left( x \right)=x\dfrac{d}{dx}{{e}^{\dfrac{1}{x}}}+{{e}^{\dfrac{1}{x}}}\dfrac{d}{dx}x \\\ & \Rightarrow f'\left( x \right)=x{{e}^{\dfrac{1}{x}}}\left( \dfrac{-1}{{{x}^{2}}} \right)+{{e}^{\dfrac{1}{x}}}-1 \\\ & \text{here }\dfrac{d}{dx}\left( {{e}^{\dfrac{1}{x}}} \right)={{e}^{\dfrac{1}{x}}}\left( \dfrac{-1}{{{x}^{2}}} \right)\text{ and }\dfrac{d}{dx}\left( x \right)=1 \\\ & \Rightarrow f'\left( x \right)=x{{e}^{\dfrac{1}{x}}}\left( \dfrac{-1}{{{x}^{2}}} \right)+{{e}^{\dfrac{1}{x}}} \\\ & \Rightarrow f'\left( x \right)=\dfrac{-{{e}^{\dfrac{1}{x}}}}{x}+{{e}^{\dfrac{1}{x}}} \\\ \end{aligned}$$ Now, using the value of $f\left( x \right)=x{{e}^{\dfrac{1}{x}}}\text{ and f }\\!\\!'\\!\\!\text{ }\left( x \right)={{e}^{\dfrac{1}{x}}}-\dfrac{{{e}^{\dfrac{1}{x}}}}{x}$ in equation (i) we get: $$I={{e}^{x}}\left\\{ f\left( x \right)+f'\left( x \right) \right\\}dx$$ Now, finally we will use a formula which states that $$\begin{aligned} & \int{{{e}^{x}}\left\\{ f\left( x \right)+f'\left( x \right) \right\\}dx}={{e}^{x}}f\left( x \right)+c \\\ & \Rightarrow I=\int{{{e}^{x}}\left\\{ {{e}^{x}}\left( {{e}^{\dfrac{1}{x}}}+x{{e}^{\dfrac{1}{x}}}-\dfrac{1}{x}{{e}^{\dfrac{1}{x}}} \right) \right\\}dx} \\\ \end{aligned}$$ Using above formula, we get: $$I={{e}^{x}}\left\\{ x{{e}^{\dfrac{1}{x}}} \right\\}+c$$ So, the value of integral $$I={{e}^{x}}\left\\{ x{{e}^{\dfrac{1}{x}}} \right\\}+c$$ $$I=x{{e}^{x}}-{{e}^{\dfrac{1}{x}}}+c$$ Taking $${{e}^{x}}\times {{e}^{\dfrac{1}{x}}}={{e}^{x+\dfrac{1}{x}}}$$ $$I=x{{e}^{x+\dfrac{1}{x}}}+c$$ Hence, the value of integral $I=x{{e}^{x+\dfrac{1}{x}}}+c$ **So, the correct answer is “Option B”.** **Note:** Another way to solve this question is by using product rule of integration, which is given as $$\int{f\left( x \right)}g\left( x \right)dx=f\left( x \right)\int{g\left( x \right)}dx-\int{\dfrac{d}{dx}\left( f\left( x \right) \right)}\int{g\left( x \right)}dxdx$$ here we can open the bracket $\left( 1+x-\dfrac{1}{x} \right){{e}^{x+\dfrac{1}{x}}}$ and take $1\times {{e}^{x+\dfrac{1}{x}}}+x{{e}^{x+\dfrac{1}{x}}}-\dfrac{1}{x}{{e}^{x+\dfrac{1}{x}}}$ and apply product rule of integration separately to get the results. Although this method is long and involves a lot of calculation, mistakes are possible. So, this method should be avoided.