Question

What is the integral of $\int{\left[ \dfrac{x{{e}^{2x}}}{{{\left( 2x+1 \right)}^{2}}} \right]}dx$ ?

Answer 1

Here in this question we have been asked to find the integral of $\int{\left[ \dfrac{x{{e}^{2x}}}{{{\left( 2x+1 \right)}^{2}}} \right]}dx$ for answering this question we will use the integration by parts concept and consider $u={{e}^{2x}}$ and $dv=\dfrac{x}{{{\left( 2x+1 \right)}^{2}}}dx$ then assume that $t=2x+1$ and simplify it further.

Complete step by step answer:
Now considering from the questions we have been asked to find the integral of $\int{\left[ \dfrac{x{{e}^{2x}}}{{{\left( 2x+1 \right)}^{2}}} \right]}dx$ .
From the basic concepts we know that the integration can be performed by splitting the integral into parts this is known as integration by parts and given as $\int{udv}=uv-\int{vdu}$ where $u$ and $v$ are functions of $v$ .
Firstly we will assume $u={{e}^{2x}}$ and $dv=\dfrac{x}{{{\left( 2x+1 \right)}^{2}}}dx$ .
Now we can say that $du=2{{e}^{2x}}dx$ because $\dfrac{d}{dx}{{e}^{ax}}=a{{e}^{ax}}$ .
Now we will simplify this further and then we will have $\int{\dfrac{x}{{{\left( 2x+1 \right)}^{2}}}dx}=v$ .
Now for further simplification we will assume $t=2x+1$ and then simplify it and have $x=\dfrac{1}{2}\left( t-1 \right)$ and then by applying derivation on both sides of it we will have $dx=\dfrac{1}{2}dt$ .
So now we can say that
$\begin{aligned} & \int{\dfrac{x}{{{\left( 2x+1 \right)}^{2}}}dx}=v \\\ & \Rightarrow \int{\dfrac{\dfrac{1}{2}\left( t-1 \right)}{{{t}^{2}}}\dfrac{1}{2}dt=\dfrac{1}{4}\int{\left( \dfrac{1}{t}-\dfrac{1}{{{t}^{2}}} \right)dt}} \\\ \end{aligned}$ .
Now by applying the formula $\int{\dfrac{1}{x}dx=\ln \left| x \right|+c}$ and $\int{{{x}^{n}}dx=}\dfrac{{{x}^{n+1}}}{n+1}+c$ where $c$ is a constant.
We will have $\dfrac{1}{4}\int{\left( \dfrac{1}{t}-\dfrac{1}{{{t}^{2}}} \right)dt}\Rightarrow \dfrac{1}{4}\left( \ln \left| t \right|+\dfrac{1}{t} \right)+C$ .
Now by substituting $t=2x+1$ we will have $\Rightarrow \dfrac{1}{4}\left( \ln \left| 2x+1 \right|+\dfrac{1}{2x+1} \right)+C$ .
Now by coming to the original question we will have $\begin{aligned} & \int{\left[ \dfrac{x{{e}^{2x}}}{{{\left( 2x+1 \right)}^{2}}} \right]}dx=\dfrac{1}{4}{{e}^{2x}}\left( \ln \left| 2x+1 \right|+\dfrac{1}{2x+1} \right)-\int{2{{e}^{2x}}}\dfrac{1}{4}\left( \ln \left| 2x+1 \right|+\dfrac{1}{2x+1} \right)dx \\\ & \Rightarrow \dfrac{{{e}^{2x}}}{4}\ln \left| 2x+1 \right|+\dfrac{{{e}^{2x}}}{4\left( 2x+1 \right)}-\dfrac{1}{2}\int{{{e}^{2x}}\ln \left| 2x+1 \right|dx-\dfrac{1}{2}\int{\dfrac{{{e}^{2x}}}{2x+1}dx}} \\\ \end{aligned}$
Now by applying integration by parts concept again on the last part we will consider $p=\dfrac{-1}{2}{{e}^{2x}}\Rightarrow dp=-{{e}^{2x}}dx$ and $dm=\dfrac{dx}{2x+1}\Rightarrow m=\dfrac{1}{2}\ln \left| 2x+1 \right|$ we will have $\begin{aligned} & \Rightarrow \dfrac{{{e}^{2x}}}{4}\ln \left| 2x+1 \right|+\dfrac{{{e}^{2x}}}{4\left( 2x+1 \right)}-\dfrac{1}{2}\int{{{e}^{2x}}\ln \left| 2x+1 \right|dx}-\dfrac{{{e}^{2x}}}{4}\ln \left| 2x+1 \right|+\dfrac{1}{2}\int{{{e}^{2x}}\ln \left| 2x+1 \right|dx} \\\ & \Rightarrow \dfrac{{{e}^{2x}}}{4\left( 2x+1 \right)}+C \\\ \end{aligned}$ .
Therefore we can conclude that the value of the integral will be given as $\int{\left[ \dfrac{x{{e}^{2x}}}{{{\left( 2x+1 \right)}^{2}}} \right]}dx=\dfrac{{{e}^{2x}}}{4\left( 2x+1 \right)}+C$ .

Note: While answering complex questions like this one our concepts should be very clear and we should make note of our steps we perform. It would be better if we practice more and more questions from this concept to get them cleared. Someone can miss one part like anyone can end up with $\int{\left[ \dfrac{x{{e}^{2x}}}{{{\left( 2x+1 \right)}^{2}}} \right]}dx=\dfrac{{{e}^{2x}}}{4}\ln \left| 2x+1 \right|+\dfrac{{{e}^{2x}}}{4\left( 2x+1 \right)}-\dfrac{1}{2}\int{{{e}^{2x}}\ln \left| 2x+1 \right|dx}-\dfrac{1}{2}\int{\dfrac{{{e}^{2x}}}{2x+1}dx}$
Because they have thought that this cannot be further simplified and messed up in between so these kinds of things get clear with practice.