Question

The value of integral \[\int{\left( 1+x-\dfrac{1}{x} \right){{e}^{x+\dfrac{1}{x}}}dx}\] is equal to
\[\begin{align}
  & A.\left( x-1 \right){{e}^{x+\dfrac{1}{x}}}+c \\
 & B.x{{e}^{x+\dfrac{1}{x}}}+c \\
 & C.\left( x+1 \right){{e}^{x+\dfrac{1}{x}}}+c \\
 & D.-x{{e}^{x+\dfrac{1}{x}}}+c \\
\end{align}\]

Answer 1

Hint: 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
\[\begin{align}
  & \dfrac{d}{dx}f\left( x \right)=f'\left( x \right)=x\dfrac{d}{dx}{{e}^{\dfrac{1}{x}}}+{{e}^{\dfrac{1}{x}}}\dfrac{d}{dx}x \\
 & \Rightarrow f'\left( x \right)=x{{e}^{\dfrac{1}{x}}}\left( \dfrac{-1}{{{x}^{2}}} \right)+{{e}^{\dfrac{1}{x}}}-1 \\
 & \text{here }\dfrac{d}{dx}\left( {{e}^{\dfrac{1}{x}}} \right)={{e}^{\dfrac{1}{x}}}\left( \dfrac{-1}{{{x}^{2}}} \right)\text{ and }\dfrac{d}{dx}\left( x \right)=1 \\
 & \Rightarrow f'\left( x \right)=x{{e}^{\dfrac{1}{x}}}\left( \dfrac{-1}{{{x}^{2}}} \right)+{{e}^{\dfrac{1}{x}}} \\
 & \Rightarrow f'\left( x \right)=\dfrac{-{{e}^{\dfrac{1}{x}}}}{x}+{{e}^{\dfrac{1}{x}}} \\
\end{align}\]
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
\[\begin{align}
  & \int{{{e}^{x}}\left\{ f\left( x \right)+f'\left( x \right) \right\}dx}={{e}^{x}}f\left( x \right)+c \\
 & \Rightarrow I=\int{{{e}^{x}}\left\{ {{e}^{x}}\left( {{e}^{\dfrac{1}{x}}}+x{{e}^{\dfrac{1}{x}}}-\dfrac{1}{x}{{e}^{\dfrac{1}{x}}} \right) \right\}dx} \\
\end{align}\]
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.