# Derivative of ln(x+2)

1. Feb 24, 2004

### JanusII

1. INT {-1 to 1} ln(x+2)dx
u=x+2
du=dx
1/(x+2)
When I use FTC, I get -.6667, not what I get with a calculator.

2. INT {2 to 3} dx/x^2-x
ln|x^2 -x|
then FTC, and different answer from calc.

I'm sure these are connected somehow. I havn't touched these in a long time, so I am guessing I am forgetting some step...

2. Feb 24, 2004

### paul11273

You are right about the connection.

1/(x+2) is the derivative of ln(x+2), not the integral. You need to go the other way. Let me know if this helps.

3. Feb 25, 2004

### JanusII

Well I tried the first one by parts.

u=ln(x+2)
du=1/(x+2) dx
dv=dx
v=x
xln(x+2) - Integral 1/(x+2)dx
u=x+2
du=dx
Int 1/u du
xln(x+2) - ln|x+2| {-1 to 1}
Not getting the correct answer I don't believe still, still doing something wrong?

4. Feb 25, 2004

### paul11273

Integration by parts is how I did it.
I am a little rusty, so bare with me...

I think you error lies in how you set up this part
You should have had ...-Integral x*1/(x+2)dx
because it is -Integral v du, and you let v=x.

I used a little substitution to simplify the (x+2) portion.
let s=x+2, this makes it a little easier to follow.

then you are solving int(ln(s)ds)
separate by parts into u=ln(s), du=1/s ds, dv=ds, v=s

Now you have s*ln(s)-int(s*(1/s) ds)
s*(1/s) = s/s = 1
so the int(s*(1/s) ds) becomes int(ds)

s*ln(s)-int(ds)
s*ln(s)-s

Now sub back (x+2) for s...
(x+2)*ln(x+2)-(x+2)