Hi. I would appreciate it if someone could show me how to integrate Ln (x+1). I have done integration by parts once arrived at xLn (x+1) minus the integration of x over (x+1).
Many thanks
Jay
I'll start from zero just so that is clearer:
$\displaystyle
\int \ln{(x+1)} dx = (\int dx) \ln{(x+1)} - \int \frac{x}{x + 1} dx
$
Now I suppose this is what you have done so far.
Try using the following equivalence ( which is obtained by dividing x by x+1):
$\displaystyle
\frac{x}{x + 1} = 1 - \frac{1}{x + 1}
$
This is a familiar integral
$\displaystyle \int lnx dx= xlnx - x + c$
Does the addition of 1 change this?
$\displaystyle U=ln(x+1)$
$\displaystyle dU = \frac{1}{x+1}$
$\displaystyle dV=dx$
$\displaystyle V=x$
$\displaystyle \int ln(x+1) dx= xln(x+1) - \int \frac{x}{x+1}dx$
Which is where you got. Let us look at the integral then.
$\displaystyle \int \frac{x}{x+1}dx = \int \frac{x+1-1}{x+1}dx = \int 1 - \frac{1}{x+1} dx$
Of course this this the same as
$\displaystyle \int 1 - \frac{1}{x+1}dx = x-lnx + c$
We then get
$\displaystyle \int ln(x+1) dx= xlnx - x + lnx + c$
Edit- LOL at how many people jumped on answering this guy's question. I should quit here there's too many of us!
Just in case a picture helps...
http://www.mathhelpforum.com/math-he...tml#post256175
... is integration by parts, where
... is the product rule.
Straight continuous lines differentiate downwards (integrate up) with respect to x. Choosing legs crossed or un-crossed does for assigning u du and v dv.
Hope it helps, or amuses - but use whatever works for you.
Cheers.
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