# How do you integrate int ln(2x+1) by integration by parts method?

Nov 8, 2016

$x \ln \left(2 x + 1\right) - x + \frac{1}{2} \ln \left\mid 2 x + 1 \right\mid + c$

#### Explanation:

$\int 1 \cdot \left(2 x + 1\right) \mathrm{dx} = x \ln \left(2 x + 1\right) - 2 \int \frac{x}{2 x + 1} \mathrm{dx}$

Substituting $y = 2 x + 1 \setminus \setminus \setminus \implies x = \frac{y - 1}{2} \setminus \setminus \setminus \implies \mathrm{dx} = \frac{1}{2} \mathrm{dy}$
$\int \frac{x}{2 x + 1} \mathrm{dx} = \int \left(y - \frac{1}{2}\right) \cdot \frac{1}{y} \cdot \frac{1}{2} \mathrm{dy} = \frac{1}{4} \int \left(1 - \frac{1}{y}\right) \mathrm{dy} = \frac{1}{4} \left(y - \ln \left\mid y \right\mid\right) + c = \frac{1}{4} \left(2 x - \ln \left\mid 2 x + 1 \right\mid\right) + c$

$\int \left(2 x + 1\right) \mathrm{dx} = x \ln \left(2 x + 1\right) - x + \frac{1}{2} \ln \left\mid 2 x + 1 \right\mid + c$

Nov 8, 2016

$\int \ln \left(2 x + 1\right) \mathrm{dx} = x \ln \left(2 x + 1\right) - x + \frac{1}{2} \ln \left(2 x + 1\right) + C$

Which can also be written:
$\int \ln \left(2 x + 1\right) \mathrm{dx} = \frac{1}{2} \left(2 x + 1\right) \ln \left(2 x + 1\right) - x + C$

#### Explanation:

If you are studying maths, then you should learn the formula for Integration By Parts (IBP), and practice how to use it:

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

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 $\ln \left(2 x + 1\right)$, you will first note that we don't actually have a product of two functions but we can use the "trick" of writing it as $\left(1\right) \ln \left(2 x + 1\right)$ hopefully now you can see that $1$ remains simple when integrated and ln(2x+1) also simplifies when differentiation.

Let $\left\{\begin{matrix}u = \ln \left(2 x + 1\right) & \implies & \frac{\mathrm{du}}{\mathrm{dx}} = \frac{2}{2 x + 1} & \text{By the chain rule} \\ \frac{\mathrm{dv}}{\mathrm{dx}} = 1 & \implies & v = x & \null\end{matrix}\right.$

Then plugging into the IBP formula givs us:
$\int \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \mathrm{dx} = \left(u\right) \left(v\right) - \int \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \mathrm{dx}$
$\therefore \int \left(\ln \left(2 x + 1\right)\right) \left(1\right) \mathrm{dx} = \left(\ln \left(2 x + 1\right)\right) \left(x\right) - \int \left(x\right) \left(\frac{2}{2 x + 1}\right) \mathrm{dx}$
$\therefore \int \ln \left(2 x + 1\right) \mathrm{dx} = x \ln \left(2 x + 1\right) - \int \frac{2 x}{2 x + 1} \mathrm{dx}$
$\therefore \int \ln \left(2 x + 1\right) \mathrm{dx} = x \ln \left(2 x + 1\right) - \int \frac{2 x}{2 x + 1} \mathrm{dx}$

So although it may not look like it we are making good progress, as we can deal with this next integral by substitution as follows:

$\left\{\begin{matrix}\text{Let" & u= 2x+1 & => (du)/dx=2 => int ... 2xdx = int ... du \\ "Then} & x = \frac{1}{2} \left(u - 1\right) & \null\end{matrix}\right.$

Substituting into the integral gives:

$\int \frac{2 x}{2 x + 1} \mathrm{dx} = \int \frac{\frac{1}{2} \left(u - 1\right)}{u} \mathrm{du}$
$\therefore \int \frac{2 x}{2 x + 1} \mathrm{dx} = \frac{1}{2} \int 1 - \frac{1}{u} \mathrm{du}$
$\therefore \int \frac{2 x}{2 x + 1} \mathrm{dx} = \frac{1}{2} \left(u - \ln u\right)$
 :. int(2x)/(2x+1)dx = 1/2(2x+1 - ln(2x+1)
$\therefore \int \frac{2 x}{2 x + 1} \mathrm{dx} = x + \frac{1}{2} - \frac{1}{2} \ln \left(2 x + 1\right)$

Combining these results we get:

$\therefore \int \ln \left(2 x + 1\right) \mathrm{dx} = x \ln \left(2 x + 1\right) - \left\{x + \frac{1}{2} - \frac{1}{2} \ln \left(2 x + 1\right)\right\} + C '$
$\therefore \int \ln \left(2 x + 1\right) \mathrm{dx} = x \ln \left(2 x + 1\right) - x - \frac{1}{2} + \frac{1}{2} \ln \left(2 x + 1\right) + C '$
$\therefore \int \ln \left(2 x + 1\right) \mathrm{dx} = x \ln \left(2 x + 1\right) - x + \frac{1}{2} \ln \left(2 x + 1\right) + C$

(NB In the last step I have eliminated the constant -1/2 by writing $C = C ' - \frac{1}{2}$)

Which can also be written as:
$\int \ln \left(2 x + 1\right) \mathrm{dx} = \frac{1}{2} \left(2 x + 1\right) \ln \left(2 x + 1\right) - x + C$