# Simple integration

1. May 10, 2008

### DeanBH

i know how to integrate.

but why is 3/x 3 log x

3x^-1 3 x^0 X^0 = 1 3*1 = 3.

i know I'm ignoring the divide by zero

why does log come into it.

4e^(1/2) im messing this one up as well.

4e^(3/2) / (3/2)

= 8/3e^(3/2) thats like as far as i get it.

somehow the answers have it at 8e^1/2

can anyone explain how that is so

Last edited: May 10, 2008
2. May 10, 2008

### arunbg

Try differentiating your answer and the correct answer and see what you get (each with respect to x of course). In which case do you get the function you started off with?

3. May 10, 2008

### DeanBH

I already know that mine is wrong and theirs is right, I just don't know how to get to their answer. >.<

k : i dont know all rules of integration apparently integral of 1/x = logx

still need help on 2nd one though

Last edited: May 10, 2008
4. May 10, 2008

### rocomath

Re-write your original post, it's very hard to interpret!

5. May 10, 2008

### HallsofIvy

Staff Emeritus
Well, obviously, your answer is wrong precisely because you are "ignoring that divide by 0".

One way to show that "log(x)" comes into it is to define log(x) as
$$\int_1^x \frac{dx}{x}$$
That is done in some textbooks.

Or you can define log(x) to be the inverse function to ex. Then if y= ln(x), x= ey. Differentiating both sides by x, 1= ey dy/dx or dy/dx= 1/ey. Since ey= x, that is dy/dx= 1/x.

Since the derivative of log(x) is 1/x, the anti-derivative (indefinite integral) of log(x) is 1/x+ C.

6. May 10, 2008

### DeanBH

this is a 5 year old exam, i assume the anti-derivatives dont come on my exam now, as the teacher hasnt taught them and they arnt on the last few year exams.

so il leave it for now. =)