Question

Find the value of \[\int {{e^{2x}}(\dfrac{1}{x} - \dfrac{1}{{2{x^2}}})dx} \] is

Answer 1

Hint: We will start with assuming \[t = 2x\] and using this we will try to find a simplified version of the given problem. We then use, the formula, \[{e^x}[f(x) + f'(x)]dx = {e^x}f(x) + c\] to get our desired result.

Complete step by step solution: We have, \[\int {{e^{2x}}(\dfrac{1}{x} - \dfrac{1}{{2{x^2}}})dx} \]
Let us take,\[t = 2x\],
\[ \Rightarrow dt = 2dx\]
\[ \Rightarrow \dfrac{{dt}}{2} = dx\]
Substituting, we get,
\[ = \dfrac{1}{2}\int {{e^t}(\dfrac{1}{{\dfrac{t}{2}}} - \dfrac{1}{{2{{(\dfrac{t}{2})}^2}}})dt} \]
On simplification we get,
\[ = \dfrac{1}{2}\int {{e^t}(\dfrac{2}{t} - \dfrac{1}{{2(\dfrac{{{t^2}}}{4})}})dt} \]
\[ = \dfrac{1}{2}\int {{e^t}(\dfrac{2}{t} - \dfrac{2}{{{t^2}}})dt} \]

On Multiplying by \[\dfrac{1}{2}\]we get,
\[ = \int {{e^t}(\dfrac{1}{t} - \dfrac{1}{{{t^2}}})dt} \]
Now, it is of the form, \[{e^x}[f(x) + f'(x)]dx = {e^x}f(x) + c\]
Where, \[f(x) = \dfrac{1}{t}\]and \[f'(x) = - \dfrac{1}{{{t^2}}}\]
So, we have now,
\[ = \dfrac{{{e^t}}}{t} + c\]
If we substitute,\[t = 2x\],
We will get,
\[ = \dfrac{{{e^{2x}}}}{{2x}} + c\]where c is the constant term.

So, our answer is, \[\int {{e^{2x}}(\dfrac{1}{x} - \dfrac{1}{{2{x^2}}})dx} \]\[ = \dfrac{{{e^{2x}}}}{{2x}} + c\]

Note: Adding constant term after the integration is very important. And also then we need to know that, \[\dfrac{d}{{dx}}(\dfrac{1}{x}) = - \dfrac{1}{{{x^2}}}\]which is used in one part of the problem. And whenever we are dealing with \[{e^x}\] we will consider \[{e^x}\]\[{\text{ = t}}\] most of the time.