# Homework Help: Reverse diferentiation problem

1. Aug 9, 2007

### cd246

1. The problem statement, all variables and given/known data

2. Relevant equations
Just reverse diferentiation

3. The attempt at a solution
=x^2+3x-1+x^-4
=(x^3)/(3)+(3x^2)/(2)-x+(1/3)(x^-3+C)
(1/3)(x^3)+(3/2)(x^2)-x+(1/3)(x^-3)+C
I know i got the answer but somehow it is backwards(exponents mostly). What am I doing wrong?

2. Aug 9, 2007

### Winzer

Look at your signs on the exponents before you intergrate and be careful, that should take care of it.

3. Aug 9, 2007

### malawi_glenn

Remember in "backwards differentiation":
i) Exponent +1
ii) divide with the new exponent.

$$(x^2+3x-1)/x^4 = x^{-2} + 3x^{-3} - x^{-4}$$

the first term:
$$x^{-2} \rightarrow (x^{-2+1})/(-2+1) = -x^{-1}$$
etc.

4. Aug 9, 2007

### HallsofIvy

Perhaps it would help if you told us what the question is!

Are you trying to integrate (x^2+ 3x-1)/x^4?

You then write "= x^2+ 3x-1+ x^-4" which is not at all the same thing: you are adding x^-4 rather than dividing by x^4.

(x^2+ 3x- 1)/x^4= (x^2/x^4)+ 3(x/x^4)- 1/x^4= x^-2+ 3x^-3- x^-4.
Now integrate.

5. Aug 9, 2007

### cd246

Now I know what I did wrong now, I just put x^-4 on top instead of subtracting the lower from the upper which I should have done. Thanks glenn