Thread: Integration of Nat. Log of (x+1)

1. Integration of Nat. Log of (x+1)

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

2. Is...

$\displaystyle \int \ln t \cdot dt = t\cdot \ln t - t + c$ (1)

Now set in (1) $\displaystyle 1+x$ instead of $\displaystyle t$ ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. Originally Posted by jaijay32
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
Split the fraction $\displaystyle \frac x{x+1} = 1-\frac1{x+1}$. The sum at the RHS can be integrated quite easily.

4. $\displaystyle \int \frac{x}{x+1} \, dx = \int \frac{x+1-1}{x+1} \, dx = \int \left( 1 - \frac{1}{x+1} \right) \, dx = \int \, dx - \int \frac{1}{x+1} \, dx$

5. Integration by parts is correct.

Originally Posted by jaijay32
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):

$\displaystyle \frac{x}{x + 1} = 1 - \frac{1}{x + 1}$

6. Originally Posted by jaijay32
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
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!

7. Originally Posted by tom@ballooncalculus
That picture ****s me up hard. Can you describe that please lol

8. ... 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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