Question

Evaluate the integral of the following
$\int{{{e}^{x}}\left( \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right)dx}$

Answer 1

Hint:If $f(x)$ is derivable function of x, then $\int{{{e}^{x}}\left[ f(x)+f'(x) \right]dx}={{e}^{x}}f(x)+c$ , where c is the constant of integration. This theorem can be used when the multiple of ${{e}^{x}}$ can be expressed as$f(x)+f'(x)$.

Complete step-by-step answer:
Let $I=\int{{{e}^{x}}\left( \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right)dx}..........(1)$
Comparing the given integral which is represented by the equation (1) with $\int{{{e}^{x}}\left[ f(x)+f'(x) \right]dx}$ , we get
The function $f(x)=\dfrac{1}{x}$ and the differentiation of the function $f(x)$ is $f'(x)=\dfrac{-1}{{{x}^{2}}}$
Let ${{e}^{x}}f(x)={{e}^{x}}\left( \dfrac{1}{x} \right)=t.........(2)$
The equation (2) is differentiating with respective to x by using product rule of differentiation and then we have
${{e}^{x}}\cdot \left( \dfrac{-1}{{{x}^{2}}} \right)+\left( \dfrac{1}{x} \right)\cdot {{e}^{x}}=\dfrac{dt}{dx}$
The arrangement of the terms, we get
${{e}^{x}}\cdot \left( \dfrac{1}{x} \right)+{{e}^{x}}\cdot \left( \dfrac{-1}{{{x}^{2}}} \right)=\dfrac{dt}{dx}$
From left hand side, we are taking the term ${{e}^{x}}$ common and then we have
${{e}^{x}}\left[ \left( \dfrac{1}{x} \right)+\left( \dfrac{-1}{{{x}^{2}}} \right) \right]=\dfrac{dt}{dx}$
Multiplying both sides by$dx$, we get
${{e}^{x}}\left[ \left( \dfrac{1}{x} \right)+\left( \dfrac{-1}{{{x}^{2}}} \right) \right]dx=dt$
The term in left hand side is also written as
${{e}^{x}}\left[ \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right]dx=dt$
Taking the integration on both sides with respective to x, we get
$\int{{{e}^{x}}\left[ \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right]dx}=t+c.........(3)$
Where c is the constant of integration.
Now, substitute the value of $t$ which is represented by the equation (2) in the equation (3), we get
$\int{{{e}^{x}}\left[ \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right]dx}={{e}^{x}}\left( \dfrac{1}{x} \right)+c$
Where c is the constant of integration.
$\int{{{e}^{x}}\left[ \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right]dx}=\dfrac{{{e}^{x}}}{x}+c$
Where c is the constant of integration.
This is the required final value of the given integral which is represented by equation (1).

Note: Alternatively, the given integral can be evaluated as follow.

& \dfrac{d}{dx}\left[ {{e}^{x}}\cdot \dfrac{1}{x} \right]={{e}^{x}}\dfrac{d}{dx}\left( \dfrac{1}{x} \right)+\dfrac{1}{x}\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}}\left( \dfrac{-1}{{{x}^{2}}} \right)+\dfrac{1}{x}{{e}^{x}} \\\ & \dfrac{d}{dx}\left[ {{e}^{x}}\cdot \dfrac{1}{x} \right]={{e}^{x}}\left[ \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right] \\\ & {{e}^{x}}\left[ \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right]=\dfrac{d}{dx}\left[ {{e}^{x}}\cdot \dfrac{1}{x} \right] \\\ \end{aligned}$$ Therefore, by the definition of indefinite integral, $\int{{{e}^{x}}\left[ \dfrac{1}{x}-\dfrac{1}{{{x}^{2}}} \right]}dx={{e}^{x}}\left( \dfrac{1}{x} \right)+c$