What is the integral of $xe^{2x}$ divided by the square of $1+2x$?

[karush]

$\tiny\text{LCC 206 {7.r.39} Integral log }$

$$\int\frac{xe^{2x}}{\left(1+2x\right)^2 }\ dx $$

Since $2x$ is in numerator and denominator thot it might be
A good candidate for

$\begin{align}\displaystyle
u& = 2x &
du&= 2 \ d{t} \\
\end{align}$

Then
$$\frac{1}{2}\int\frac{ue^{u}}{\left(1+u\right)^2 }\ dx $$
But?

$\tiny\text
{from Surf the Nations math study group}$

 

[MarkFL]

I would look at IBP here...let:

\(\displaystyle u=xe^{2x}\)

and

\(\displaystyle dv=(2x+1)^{-2}\,dx\)

What do you get?

 

[MarkFL]

How do you isolate $x$

What do you mean exactly? Why do you need to isolate $x$ in order to find $du$ and $v$?

 

[karush]

$\displaystyle
\begin{align}
u& = xe^{2x}&
dv&=\frac{1}{\left(2x+1\right)^2 } \ d{x} \\
du& = \left(2x+1\right)e^{2x}&
v&=\frac{-1}{2\left(2x+1\right) } \\
\end{align}$

$\displaystyle
\frac{-xe^{2x}}{2\left(2x+1\right)}
+\frac{1}{2}\int\frac{\left(2x+1\right)e^{2x}}
{\left(2x+1\right) } \ dx
\implies
\frac{-xe^{2x}}{2\left(2x+1\right)}
+\frac{1}{2}e^{2x}+C$

 

[MarkFL]

$\displaystyle
\begin{align}
u& = xe^{2x}&
dv&=\frac{1}{\left(2x+1\right)^2 } \ d{x} \\
du& = \left(2x+1\right)e^{2x}&
v&=\frac{-1}{2\left(2x+1\right) } \\
\end{align}$

$\displaystyle
\frac{-xe^{2x}}{2\left(2x+1\right)}
+\frac{1}{2}\int\frac{\left(2x+1\right)e^{2x}}
{\left(2x+1\right) } \ dx $

Okay, good, now simplify the remaining integral, and you can integrate it directly to get your final result. :D

 

[karush]

$\displaystyle
\frac{-xe^{2x}}{2\left(2x+1\right)}
+\frac{1}{2}\int\frac{\left(2x+1\right)e^{2x}}
{\left(2x+1\right) } \ dx
\implies
\frac{-xe^{2x}}{2\left(2x+1\right)}
+\frac{1}{2}e^{2x}+C$

 

[MarkFL]

$\displaystyle
\frac{-xe^{2x}}{2\left(2x+1\right)}
+\frac{1}{2}\int\frac{\left(2x+1\right)e^{2x}}
{\left(2x+1\right) } \ dx
\implies
\frac{-xe^{2x}}{2\left(2x+1\right)}
+\frac{1}{2}e^{2x}+C$

Check your integration...you are missing a constant factor because you are integrating $e^{2x}$ rather than $e^x$. :)

I may have misled you when I said it could be "integrated directly"...it technically requires a substitution, but this can be done mentally. :D