$\int {(1 + x - \dfrac{1}{x}){e^{x + \dfrac{1}{x}}}dx} $ is equal to
A. $(x - 1){e^{x + \dfrac{1}{x}}} + c$
B. $x{e^{x + \dfrac{1}{x}}} + c$
C. $(x + 1){e^{x + \dfrac{1}{x}}} + c$
D. $ - x{e^{x + \dfrac{1}{x}}} + c$

Hint: In order to solve the question above, we will be using the formula of integration by parts. It is used when a product of two functions are given to be integrated. If the two functions are $u$ and $v$, then the formula is stated as $\int {uvdx = u\int {vdx} - \int {(\dfrac{{du}}{{dx}}\int {vdx)dx} } } $.

Complete step by step solution:
The given expression in the question is $\int {(1 + x - \dfrac{1}{x}){e^{x + \dfrac{1}{x}}}dx} $.
We will break the given expression into two expressions such that it becomes
$ \Rightarrow \int {(1 + x - \dfrac{1}{x}){e^{x + \dfrac{1}{x}}}dx} = \int {{e^{x + \dfrac{1}{x}}}dx + \int {(x - \dfrac{1}{x})} } {e^{x + \dfrac{1}{x}}}dx$ {equation (1)}
Let the second term $\int {(x - \dfrac{1}{x}){e^{x + \dfrac{1}{x}}}dx} $ on the right hand side of the given equation be named as $A$ such that
$A = \int {(x - \dfrac{1}{x}){e^{x + \dfrac{1}{x}}}dx} $ {equation (2)}
On substituting equation (2) in equation (1), we will get
\[ \Rightarrow \int {(1 + x - \dfrac{1}{x}){e^{x + \dfrac{1}{x}}}dx} = \int {{e^{x + \dfrac{1}{x}}}dx + A} \] {equation (3)
Moreover, the expression A can further be written as
$ \Rightarrow A = \int {(x - \dfrac{1}{x}){e^{x + \dfrac{1}{x}}}dx} $
$ \Rightarrow A = \int {x(1 - \dfrac{1}{{{x^2}}}){e^{x + \dfrac{1}{x}}}dx} $
We will now apply the formula for integration by-parts which is given as $\int {uvdx = u\int {vdx} - \int {(\dfrac{{du}}{{dx}}\int {vdx)dx} } } $.
Here let $u = x$ and $v = (1 - \dfrac{1}{{{x^2}}}){e^{x + \dfrac{1}{x}}}$.
On putting the respective values of $u$ and $v$ in the by-parts formula, we will get
$ \Rightarrow \int {uvdx = u\int {vdx} - \int {(\dfrac{{du}}{{dx}}\int {vdx)dx} } } $
$ \Rightarrow \int {x(1 - \dfrac{1}{{{x^2}}}){e^{x + \dfrac{1}{x}}}dx = x\int {(1 - \dfrac{1}{{{x^2}}}){e^{x + \dfrac{1}{x}}}dx} - \int {(\dfrac{{dx}}{{dx}}\int {(1 - \dfrac{1}{{{x^2}}}){e^{x + \dfrac{1}{x}}}dx)dx} } } $
In the above step, let the expression $\int {(1 - \dfrac{1}{{{x^2}}}){e^{x + \dfrac{1}{x}}}dx} $ be $B$, such that the above given equation becomes
$ \Rightarrow A = xB - \int {(\dfrac{{dx}}{{dx}}B)dx} $ {equation (4)}
Now we will firstly solve $B$. We have
$B = \int {(1 - \dfrac{1}{{{x^2}}}){e^{x + \dfrac{1}{x}}}dx} $ {equation (5)}
In the above expression let $x + \dfrac{1}{x} = t$ {equation (6)}
When we differentiate both the sides, we get
$ \Rightarrow (1 - \dfrac{1}{{{x^2}}})dx = dt$ {equation (7)}
On putting equation (6) and equation (7) in equation (5), we will get
$ \Rightarrow B = \int {{e^t}dt} $
We know that $\int {{e^x}dx} = {e^x}$. Hence on integrating above, we will get
$ \Rightarrow B = {e^t}$
On putting the value of $t$ from equation (6) in $B$ above, we will get
$ \Rightarrow B = {e^{x + \dfrac{1}{x}}}$
Now we will substitute this obtained value of $B$ in equation (4), such that it becomes
$ \Rightarrow A = x{e^{x + \dfrac{1}{x}}} - \int {(\dfrac{{dx}}{{dx}}{e^{x + \dfrac{1}{x}}})dx} $
We know that differentiation of $x$ with respect to $x$ is $1$, i.e. $\dfrac{{dx}}{{dx}} = 1$. So putting this above, we get
$ \Rightarrow A = x{e^{x + \dfrac{1}{x}}} - \int {{e^{x + \dfrac{1}{x}}}dx} + c$
Now putting this value of $A$ in equation (3), we will get
\[ \Rightarrow \int {(1 + x - \dfrac{1}{x}){e^{x + \dfrac{1}{x}}}dx} = \int {{e^{x + \dfrac{1}{x}}}dx + x{e^{x + \dfrac{1}{x}}} - \int {{e^{x + \dfrac{1}{x}}}dx} + c} \]
The term \[\int {{e^{x + \dfrac{1}{x}}}dx} \] is present in the above equation with both the positive and negative signs. So they cancel each other and therefore we get
\[ \Rightarrow \int {(1 + x - \dfrac{1}{x}){e^{x + \dfrac{1}{x}}}dx} = x{e^{x + \dfrac{1}{x}}} + c\]

Therefore the correct answer is option B.

Note:
It becomes sometimes confusing to choose which function as $u$ and which one as $v$. For this, we can remember the acronym ILATE where I is for inverse functions, L is for logarithmic functions, A is for algebraic functions, T is for trigonometric functions and E is for exponential functions. Suppose if we are given with a logarithmic function and an exponential function, then we must choose the former as $u$ and the latter as $v$.