Integration urgent help please.

I have -2x+4-3/2x
Integrated: -x^2 + 4x -3/2
I am not sure about the -3/2 because -3/2x gives -3x^-1/2. So -1+1 = 0 and anything to the power of 0 = 1. So that results in 3/2. Is this correct though?


Thank you in advance

integration of 1/x is lnx .

Sorry, what is Inx?

Np. lnx is the natural logarithm of x. I guess Wikipedia does a better job of explaining this than me: https://en.wikipedia.org/wiki/Natural_logarithm :P

This is because the derivative of ln x is 1/x.

Ahh...this is c2 stuff and I haven't done logs yet. So in order to integrate the function I must do logs first?

If you are working with only the algebraic problems (and some trigonometric functions involving only the sin and cos) then you should be fine by just noting that the integration of 1/x is ln x. Otherwise, you might wanna study logarithms first.

maybe without logs you could bring the 1/2x to the top by calling it 2x^-1 then integrate that

maybe without logs you could bring the 1/2x to the top by calling it 2x^-1 then integrate that

I have -2x+4-3/2x
Integrated: -x^2 + 4x -3/2
I am not sure about the -3/2 because -3/2x gives -3x^-1/2. So -1+1 = 0 and anything to the power of 0 = 1. So that results in 3/2. Is this correct though?


Thank you in advance

Do you mean -3/(2x) or (-3/2)x ?

This thread...


OP, what is your exam board? If they do not cover the natural logarithm and its derivative in C2 then you probably have the wrong question or written it down incorrectly.