Question

Integrate the following:
$\int {{e^{2x}}} \left( {\dfrac{{2x - 1}}{{4{x^2}}}} \right)dx$

Answer 1

Hint: To simplify the integration, first assume $2x = t$ and convert the integration in the form of variable $t$. Next, separate the integration into two different integrations by separating fractions. And then integrate the first one using integration by parts and the second one will get canceled giving is the answer.

Complete step-by-step solution:
According to the question, we have to integrate the given function. This integration is $\int {{e^{2x}}} \left( {\dfrac{{2x - 1}}{{4{x^2}}}} \right)dx$.
Let this integration be denoted as $I$. So we have:
$ \Rightarrow I = \int {{e^{2x}}} \left( {\dfrac{{2x - 1}}{{4{x^2}}}} \right)dx$
To integrate this function, first assume $2x = t$.
If we differentiate this both sides, we have:
$
   \Rightarrow 2dx = dt \\
   \Rightarrow dx = \dfrac{{dt}}{2} \\
 $
Substituting these values in the integration, we’ll get:
 $ \Rightarrow I = \int {{e^t}} \left( {\dfrac{{t - 1}}{{{t^2}}}} \right)\dfrac{{dt}}{2}$
We can take $\dfrac{1}{2}$ outside of the integration as it is a constant. Further, if we separate the fraction $\left( {\dfrac{{t - 1}}{{{t^2}}}} \right)$ into two different fractions, we will get:
$ \Rightarrow I = \dfrac{1}{2}\int {{e^t}} \left( {\dfrac{t}{{{t^2}}} - \dfrac{1}{{{t^2}}}} \right)dt$
On further simplification, this will give us:
$ \Rightarrow I = \dfrac{1}{2}\int {{e^t}} \left( {\dfrac{1}{t} - \dfrac{1}{{{t^2}}}} \right)dt$
Now, we can write this as two separate integrations. So we have:
$ \Rightarrow I = \dfrac{1}{2}\int {\dfrac{{{e^t}}}{t}} dt - \dfrac{1}{2}\int {\dfrac{{{e^t}}}{{{t^2}}}} dt$
In this, we will integrate the first one by parts taking $\dfrac{1}{t}$ as first function and ${e^t}$ as second function. To refresh our memory, the rule of integration by parts is shown below:
$ \Rightarrow \int {uvdx = u\int {vdx} } - \int {\dfrac{{du}}{{dx}}} \left( {\int {vdx} } \right)dx$, here $u$ and $v$ are the functions of $x$.
Using this rule for our integration as discussed above, we’ll get:
$ \Rightarrow I = \dfrac{1}{2}\left[ {\dfrac{1}{t}\int {{e^t}} dt - \int {\dfrac{d}{{dt}}\left( {\dfrac{1}{t}} \right)\left( {\int {{e^t}dt} } \right)dt} } \right] - \dfrac{1}{2}\int {\dfrac{{{e^t}}}{{{t^2}}}} dt$
We know that the differentiation of $\dfrac{1}{x}$ is $ - \dfrac{1}{{{x^2}}}$. So we have:
$ \Rightarrow \dfrac{d}{{dt}}\left( {\dfrac{1}{t}} \right) = - \dfrac{1}{{{t^2}}}$
And we also know that the integration of ${e^t}$ is ${e^t}$ itself. Using these results in our integration, we’ll get:
$ \Rightarrow I = \dfrac{1}{2}\left[ {\dfrac{{{e^t}}}{t} - \int {\left( { - \dfrac{1}{{{t^2}}}} \right){e^t}dt} } \right] - \dfrac{1}{2}\int {\dfrac{{{e^t}}}{{{t^2}}}} dt + C$
$C$ is the integration constant.
The integration can be further simplified as:
$
   \Rightarrow I = \dfrac{1}{2}\left[ {\dfrac{{{e^t}}}{t} + \int {\dfrac{{{e^t}}}{{{t^2}}}dt} } \right] - \dfrac{1}{2}\int {\dfrac{{{e^t}}}{{{t^2}}}} dt + C \\
   \Rightarrow I = \dfrac{{{e^t}}}{{2t}} + \dfrac{1}{2}\int {\dfrac{{{e^t}}}{{{t^2}}}dt} - \dfrac{1}{2}\int {\dfrac{{{e^t}}}{{{t^2}}}} dt + C \\
 $
The two integration left are the same but with opposite signs. So they will cancel out each other. We will get:
$ \Rightarrow I = \dfrac{{{e^t}}}{{2t}} + C$
Putting back the value of $t$ as $2x$, we have:
$
   \Rightarrow I = \dfrac{{{e^{2x}}}}{{2\left( {2x} \right)}} + C \\
   \Rightarrow I = \dfrac{{{e^{2x}}}}{{4x}} + C \\
 $

Putting back the integration $I$:
$ \Rightarrow \int {{e^{2x}}} \left( {\dfrac{{2x - 1}}{{4{x^2}}}} \right)dx = \dfrac{{{e^{2x}}}}{{4x}} + C$, where $C$ is integration constant.

Note: Whenever we have to integrate the multiple of two functions which cannot be integrated together while one of them can be integrated separately then we can apply integration by parts. The rule of integration by parts is given below:
$ \Rightarrow \int {uvdx = u\int {vdx} } - \int {\dfrac{{du}}{{dx}}} \left( {\int {vdx} } \right)dx$, here $u$ and $v$ are the functions of $x$.
Here, we have assumed functions $u$ and $v$ cannot be integrated together. So we used integration by parts. However the function $v$ can be easily integrated which is why this method is helpful.
But this method has its limitations for which we can’t use this in all scenarios. Sometimes the differentiation of the first function in the rule makes the integration even more complicated than what it was.