Understanding Integration: Simplifying Logarithms and Exponents

[DeanBH]

i know how to integrate.

but why is 3/x 3 log x3x^-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) I am messing this one up as well.4e^(3/2) / (3/2)

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

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

can anyone explain how that is so

 

[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?

 

[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 don't know all rules of integration apparently integral of 1/x = logx

still need help on 2nd one though

 

[rocomath]

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

 

[HallsofIvy]

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 e^x. Then if y= ln(x), x= e^y. Differentiating both sides by x, 1= e^y dy/dx or dy/dx= 1/e^y. Since e^y= 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.

 

[DeanBH]

this is a 5 year old exam, i assume the anti-derivatives don't 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. =)