Integrals

[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...

[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.

[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?

[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
xln(x+2) - Integral 1/(x+2)dx
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)

Like I said, I am a little rusty, but I feel good about this.
If anyone sees an error in my method, please let me know.
Also, if there is an easier way to go about this problem, that would be great to here too.