# Integrate int (xe^(2x))/(2x+1)^2 dx. I'm having difficulty integration it second time(integration by parts)?

Feb 5, 2017

$\int \frac{x {e}^{2 x}}{2 x + 1} ^ 2 \setminus \mathrm{dx} = {e}^{2 x} / \left(8 x + 4\right) + C$

#### Explanation:

Let $I = \int \frac{x {e}^{2 x}}{2 x + 1} ^ 2 \setminus \mathrm{dx}$

We can simplify the integral using the substitution:

$t = 2 x + 1 \implies \frac{\mathrm{dt}}{\mathrm{dx}} = 2 \setminus \setminus$ and $\setminus \setminus 2 x = t - 1$

Then:

$I = \int \setminus \frac{\frac{t - 1}{2} {e}^{t - 1}}{{t}^{2}} \setminus \left(\frac{1}{2}\right) \setminus \mathrm{dt}$
$\setminus \setminus = \frac{1}{4} \int \setminus \frac{\left(t - 1\right) {e}^{t} {e}^{-} 1}{{t}^{2}} \setminus \mathrm{dt}$
$\setminus \setminus = \frac{1}{4 e} \int \setminus {e}^{t} / t - {e}^{t} / {t}^{2} \setminus \mathrm{dt}$
$\setminus \setminus = \frac{1}{4 e} \int \setminus {e}^{t} / t \setminus \mathrm{dt} - \frac{1}{4 e} \int \setminus {e}^{t} / {t}^{2} \setminus \mathrm{dt}$ ..... [1]

Consider the second integral $\int \setminus {e}^{t} / {t}^{2} \setminus \mathrm{dt}$. we can evaluate this using integration by parts:

$\int u \frac{\mathrm{dv}}{\mathrm{dx}} \mathrm{dx} = u v - \int v \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$, or less formally $\text{ } \int u \mathrm{dv} = u v - \int v \mathrm{du}$

I was taught to remember the less formal rule in word; "The integral of udv equals uv minus the integral of vdu". If you struggle to remember the rule, then it may help to see that it comes a s a direct consequence of integrating the Product Rule for differentiation.

Essentially we would like to identify one function that simplifies when differentiated, and identify one that simplifies when integrated (or is at least is integrable).

So for the integrand ${e}^{t} / {t}^{2}$, hopefully you can see that $\frac{1}{t} ^ 2$ simplifies when integrated and ${e}^{t}$ remains unchanged.

Let $\left\{\begin{matrix}u & = {e}^{t} & \implies & \frac{\mathrm{du}}{\mathrm{dt}} & = {e}^{t} \\ \frac{\mathrm{dv}}{\mathrm{dt}} & = \frac{1}{t} ^ 2 & \implies & v & = - \frac{1}{t}\end{matrix}\right.$

Then plugging into the IBP formula gives us:

$\therefore \setminus \int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dt}}\right) \mathrm{dt} = \left(u\right) \left(v\right) - \int \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dt}}\right) \mathrm{dt}$
$\therefore \int \setminus \left({e}^{t}\right) \left(\frac{1}{t} ^ 2\right) \setminus \mathrm{dt} = \left({e}^{t}\right) \left(- \frac{1}{t}\right) - \int \left(- \frac{1}{t}\right) \left({e}^{t}\right) \setminus \mathrm{dt}$
$\therefore \int \setminus {e}^{t} / {t}^{2} \setminus \mathrm{dt} = - {e}^{t} / t + \int \setminus {e}^{t} / t \setminus \mathrm{dt} + {C}_{1}$

Substituting this into [1] gives us:

$I = \frac{1}{4 e} \int \setminus {e}^{t} / t \setminus \mathrm{dt} - \frac{1}{4 e} \left\{- {e}^{t} / t + \int \setminus {e}^{t} / t \setminus \mathrm{dt} + {C}_{1}\right\}$
$\setminus \setminus = \frac{1}{4 e} \int \setminus {e}^{t} / t \setminus \mathrm{dt} + \frac{1}{4 e} {e}^{t} / t - \frac{1}{4 e} \int \setminus {e}^{t} / t \setminus \mathrm{dt} + C$

The first and last term cancel to give;

$I = \frac{1}{4 e} {e}^{t} / t + C$

If we restore our substitution we get:

$I = {e}^{-} \frac{1}{4} \cdot {e}^{2 x + 1} / \left(2 x + 1\right) + C$
$\setminus \setminus = \frac{1}{4} \cdot {e}^{2 x} / \left(2 x + 1\right) + C$
$\setminus \setminus = {e}^{2 x} / \left(8 x + 4\right) + C$

Hence:

$\int \frac{x {e}^{2 x}}{2 x + 1} ^ 2 \setminus \mathrm{dx} = {e}^{2 x} / \left(8 x + 4\right) + C$