Question

The value of $\int {{e^{2x}}\left( {\dfrac{1}{x} - \dfrac{1}{{2{x^2}}}} \right)dx}$ is

Answer 1

We will substitute $2x = t$ and will simplify the expression. Then, we will use the identity $\int {{e^x}\left[ {f\left( x \right) + f'\left( x \right)} \right]dx = {e^x}f\left( x \right) + C}$ to solve the integral. At last, substitute back the value of $t$ to get the final answer.

Complete step-by-step answer:
We have to evaluate the value of $\int {{e^{2x}}\left( {\dfrac{1}{x} - \dfrac{1}{{2{x^2}}}} \right)dx}$.
We will solve the required integral by substitution.
Let $2x = t$, then $dx = \dfrac{{dt}}{2}$
Therefore, the given integral becomes,
$\Rightarrow \int {{e^t}\left( {\dfrac{1}{{\dfrac{t}{2}}} - \dfrac{1}{{2{{\left( {\dfrac{t}{2}} \right)}^2}}}} \right)\dfrac{{dt}}{2}} \\\ \Rightarrow \int {{e^t}\left( {\dfrac{2}{t} - \dfrac{4}{{2{t^2}}}} \right)\dfrac{{dt}}{2}} \\\ \Rightarrow \int {{e^t}\left( {\dfrac{1}{t} - \dfrac{2}{{{t^2}}}} \right)dt} \\\$
Since, $\dfrac{d}{{dt}}\left( {\dfrac{1}{t}} \right) = - \dfrac{1}{{{t^2}}}$
The given integral is of the form $\int {{e^x}\left[ {f\left( x \right) + f'\left( x \right)} \right]dx = {e^x}f\left( x \right) + C}$, where $f\left( x \right) = \dfrac{1}{t}$ and $f'\left( x \right) = - \dfrac{1}{{{t^2}}}$
Then, the value of the given integral is $\dfrac{{{e^t}}}{t} + C$
Now, put back the value of $t = 2x$
$\Rightarrow \dfrac{{{e^{2x}}}}{{2x}} + C$

Note: The function of the type ${e^x}$ is an exponential function. The integral involving ${e^x}$ is usually solved using by parts or the identity $\int {{e^x}\left[ {f\left( x \right) + f'\left( x \right)} \right]dx = {e^x}f\left( x \right) + C}$. Since, limits are not given, it is an indefinite integral.