1. ## Intergration by part

How do you do this please?

2. Write this as,
$\int (1)\ln (2x+1) \ dx$
Let $u=1$ and $v' = \ln (2x+1)$

3. Can you be more detail..please...

4. Originally Posted by camherokid
How do you do this please?
$\int \ln (2x + 1) ~dx$

Let $u = 2x + 1$

$\Rightarrow du = 2 ~dx$

$\Rightarrow \frac {1}{2} du = dx$

So our integral becomes:

$\frac {1}{2} \int \ln u ~du$

We proceed by integration by parts:

$\Rightarrow \frac {1}{2} \int \ln u ~du = \frac {1}{2} \left[u \ln u - \int u \left( \frac {1}{u} \right)~du \right]$

$= \frac {1}{2} \left[ u \ln u - \int 1 ~du + C \right]$

$= \frac {1}{2} \left[ u \ln u - u \right] + C$

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

Recall that the integration by parts formula is:

$\int uv' = uv - \int u'v$

where u and v are functions of the same variable

above i chose $u = \ln u$ and $v' = 1$

So i got $u' = \frac {1}{u}$ and $v = u$ and i just plugged that into the formula ........seeing that last line i realize i should have used a different variable for the first substituion i did, say $t$. you can redo the problem making those corrections. i don't think about the formula when i am doing by parts that's why that happened, but we can't have the variables of the formula being the same as the variables we are integrating with respect to, that causes confusion

5. Originally Posted by camherokid
How do you do this please?
$
I(x)=\int (1)\ln (2x+1) \ dx
$

let $dv=1$ and $u=\ln(2x+1)$ then by integration by parts:

$
I(x)=uv - \int v du = x \ln(2x+1) - \int \frac{2x}{2x+1}dx
$

RonL

6. Hello, camherokid!

$\int \ln(2x+1)\,dx$
There aren't too many choices, are there?

. . $\begin{array}{ccccccc}u & = & \ln(2x+1) & \quad & dv & = & dx \\ du & = & \frac{2}{2x+1}\,dx & \quad & v & = & x\end{array}$

Then we have: . $x\ln(2x+1) - \int\frac{2x}{2x+1}\,dx \;\;=\;\;x\ln(2x+1) - \int\left[1 - \frac{1}{2x+1}\right]\,dx$

. . $= \;\;x\ln(2x+1)\;-\;\left[x - \frac{1}{2}\ln(2x+1)\right]\;+\;C \;\;=\;\;x\ln(2x+1)\;-\;x\;+\;\frac{1}{2}\ln(2x+1)\;+\;C$

. . $=\;\;\left(x \:+ \:\frac{1}{2}\right)\ln(2x+1) - x\:+\:C \;\;=\;\;\frac{1}{2}(2x+1)\!\cdot\ln(2x+1)\:-\:x\:+\:C

$