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

Find the value of ∫e2x(1x−12x2)dx\int {{e^{2x}}(\dfrac{1}{x} - \dfrac{1}{{2{x^2}}})dx} is

Explanation

Solution

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

Complete step by step solution: We have, ∫e2x(1x−12x2)dx\int {{e^{2x}}(\dfrac{1}{x} - \dfrac{1}{{2{x^2}}})dx}
Let us take,t=2xt = 2x,
⇒dt=2dx\Rightarrow dt = 2dx
⇒dt2=dx\Rightarrow \dfrac{{dt}}{2} = dx
Substituting, we get,
=12∫et(1t2−12(t2)2)dt= \dfrac{1}{2}\int {{e^t}(\dfrac{1}{{\dfrac{t}{2}}} - \dfrac{1}{{2{{(\dfrac{t}{2})}^2}}})dt}
On simplification we get,
=12∫et(2t−12(t24))dt= \dfrac{1}{2}\int {{e^t}(\dfrac{2}{t} - \dfrac{1}{{2(\dfrac{{{t^2}}}{4})}})dt}
=12∫et(2t−2t2)dt= \dfrac{1}{2}\int {{e^t}(\dfrac{2}{t} - \dfrac{2}{{{t^2}}})dt}

On Multiplying by 12\dfrac{1}{2}we get,
=∫et(1t−1t2)dt= \int {{e^t}(\dfrac{1}{t} - \dfrac{1}{{{t^2}}})dt}
Now, it is of the form, ex[f(x)+f′(x)]dx=exf(x)+c{e^x}[f(x) + f'(x)]dx = {e^x}f(x) + c
Where, f(x)=1tf(x) = \dfrac{1}{t}and f′(x)=−1t2f'(x) = - \dfrac{1}{{{t^2}}}
So, we have now,
=ett+c= \dfrac{{{e^t}}}{t} + c
If we substitute,t=2xt = 2x,
We will get,
=e2x2x+c= \dfrac{{{e^{2x}}}}{{2x}} + cwhere 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, ddx(1x)=−1x2\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 ex{e^x} we will consider {e^x}$$$${\text{ = t}} most of the time.