How do you do this please?
$\displaystyle \int \ln (2x + 1) ~dx$
Let $\displaystyle u = 2x + 1$
$\displaystyle \Rightarrow du = 2 ~dx$
$\displaystyle \Rightarrow \frac {1}{2} du = dx$
So our integral becomes:
$\displaystyle \frac {1}{2} \int \ln u ~du$
We proceed by integration by parts:
$\displaystyle \Rightarrow \frac {1}{2} \int \ln u ~du = \frac {1}{2} \left[u \ln u - \int u \left( \frac {1}{u} \right)~du \right] $
$\displaystyle = \frac {1}{2} \left[ u \ln u - \int 1 ~du + C \right] $
$\displaystyle = \frac {1}{2} \left[ u \ln u - u \right] + C$
$\displaystyle = \frac {1}{2} \left( (2x + 1) \ln |2x + 1| - (2x + 1) \right) + C$
Recall that the integration by parts formula is:
$\displaystyle \int uv' = uv - \int u'v$
where u and v are functions of the same variable
above i chose $\displaystyle u = \ln u$ and $\displaystyle v' = 1$
So i got $\displaystyle u' = \frac {1}{u}$ and $\displaystyle v = u$ 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 $\displaystyle t$. 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
Hello, camherokid!
There aren't too many choices, are there?$\displaystyle \int \ln(2x+1)\,dx$
. . $\displaystyle \begin{array}{ccccccc}u & = & \ln(2x+1) & \quad & dv & = & dx \\ du & = & \frac{2}{2x+1}\,dx & \quad & v & = & x\end{array}$
Then we have: .$\displaystyle x\ln(2x+1) - \int\frac{2x}{2x+1}\,dx \;\;=\;\;x\ln(2x+1) - \int\left[1 - \frac{1}{2x+1}\right]\,dx$
. . $\displaystyle = \;\;x\ln(2x+1)\;-\;\left[x - \frac{1}{2}\ln(2x+1)\right]\;+\;C \;\;=\;\;x\ln(2x+1)\;-\;x\;+\;\frac{1}{2}\ln(2x+1)\;+\;C$
. . $\displaystyle =\;\;\left(x \:+ \:\frac{1}{2}\right)\ln(2x+1) - x\:+\:C \;\;=\;\;\frac{1}{2}(2x+1)\!\cdot\ln(2x+1)\:-\:x\:+\:C
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