Integrate \[\int {{e^x}\left( {\dfrac{1}{x} - \dfrac{1}{{{x^2}}}} \right)dx} \]

Hint:
We will denote the function inside the parenthesis as a function of \[x\]. We will then differentiate the assumed function such that its derivative is present in the integrand. Then we will apply the appropriate integration formula to obtain the value of the integral.

Formula used:
We will use the following formulas:
1) \[\int {{e^x}[f(x) + f'(x)} ]dx = {e^x}f(x) + c\], where \[f(x)\] is a function of \[x\], \[f'(x)\] is its derivative, and \[c\] is a constant.
2) \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]

Complete step by step solution:
We have to integrate \[\int {{e^x}\left( {\dfrac{1}{x} - \dfrac{1}{{{x^2}}}} \right)dx} \]
We will first use an appropriate function of \[x\] from the integrand, such that its derivative is also present in the integrand.
 Let us take \[\dfrac{1}{x}\] as our required function. So,
\[f(x) = \dfrac{1}{x}\]
Let us check if the derivative of \[\dfrac{1}{x}\] is present in the integrand. For this we have to differentiate \[\dfrac{1}{x}\]. Now, \[\dfrac{1}{x} = {x^{ - 1}}\].
Using the formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\], we get
\[\begin{array}{l}f'(x) = ( - 1){x^{ - 1 - 1}} = - {x^{ - 2}}\\ \Rightarrow f'(x) = - \dfrac{1}{{{x^2}}}\end{array}\]
We observe that \[ - \dfrac{1}{{{x^2}}}\] is also present in the integrand. Hence, our choice of function is right.
Now, we can write \[\int {{e^x}\left( {\dfrac{1}{x} - \dfrac{1}{{{x^2}}}} \right)dx} \] as \[\int {{e^x}\left[ {\dfrac{1}{x} + \left( { - \dfrac{1}{{{x^2}}}} \right)} \right]} dx\], where \[f(x) = \dfrac{1}{x}\] and \[f'(x) = - \dfrac{1}{{{x^2}}}\].
\[\int {{e^x}\left( {\dfrac{1}{x} - \dfrac{1}{{{x^2}}}} \right)dx} = \int {{e^x}\left[ {\dfrac{1}{x} + \left( { - \dfrac{1}{{{x^2}}}} \right)} \right]} dx\]
Using the formula \[\int {{e^x}[f(x) + f'(x)} ]dx = {e^x}f(x) + c\] to integrate, we have

\[ \Rightarrow \int {{e^x}\left( {\dfrac{1}{x} - \dfrac{1}{{{x^2}}}} \right)dx} = {e^x}\left( {\dfrac{1}{x}} \right) + c\]

Note:
We can also solve the above problem by integration by parts method.
\[\int {udv} = uv - \int {vdu} \] ……….\[(A)\]
The integrand \[\int {{e^x}\left( {\dfrac{1}{x} - \dfrac{1}{{{x^2}}}} \right)dx} \] can be written as \[\int {{e^x}\left( {\dfrac{1}{x}} \right)dx - \int {{e^x}\left( {\dfrac{1}{{{x^2}}}} \right)dx} } \]……….\[(B)\]
We will apply the integration by parts rule to the integral \[\int {{e^x}\left( {\dfrac{1}{x}} \right)dx} \].
Let us take \[u = \dfrac{1}{x}\] and \[dv = {e^x}dx\].
Now, \[du = - \dfrac{1}{{{x^2}}}dx\] and \[v = \int {dv} = \int {{e^x}dx} \]. This gives \[v = {e^x}\].
 Substituting these values in equation \[(A)\], we get
\[\int {{e^x}\left( {\dfrac{1}{x}} \right)dx} = {e^x}\left( {\dfrac{1}{x}} \right) - \int {{e^x}\left( { - \dfrac{1}{{{x^2}}}} \right)dx} + c\]
On the RHS, the \[( - )\] can be taken out and we can rewrite the above equation as
\[\int {{e^x}\left( {\dfrac{1}{x}} \right)dx} = {e^x}\left( {\dfrac{1}{x}} \right) + \int {{e^x}\left( {\dfrac{1}{{{x^2}}}} \right)dx} \] ………\[(C)\]
Using equation \[(C)\] in equation \[(B)\], we get
\[\int {{e^x}\left( {\dfrac{1}{x} - \dfrac{1}{{{x^2}}}} \right)dx} = {e^x}\left( {\dfrac{1}{x}} \right) + \int {{e^x}\left( {\dfrac{1}{{{x^2}}}} \right)dx} - \int {{e^x}\left( {\dfrac{1}{{{x^2}}}} \right)dx} + c\]
Cancelling out \[\int {{e^x}\left( {\dfrac{1}{{{x^2}}}} \right)dx} \] on the RHS, we finally get
\[\int {{e^x}\left( {\dfrac{1}{x} - \dfrac{1}{{{x^2}}}} \right)dx} = {e^x}\left( {\dfrac{1}{x}} \right) + c\]