# Thread: My Own Proof of uniqueness of antiderivative

1. ## (Fixed) My Own Proof that Antiderivative NOT unique

Greetings everyone. I'm trying to do the following proof my own way but it doesn't look like my final answer is matching the solution's. Is there something wrong?
The solution shows the result by picking specific numbers of $a, b, c$ and doing some tricky algebra, right? But I want to avoid the tricky algebra
so I'm trying to use long division to "pull out" or derive the integration constant.

Thank you everyone.

$\text{Given that P(}x)=Q(x)R'(x)-Q'(x)R(x),\text{ write down an expression for } \int \frac{P}{Q^2} dx.$
$\text{By choosing } R(x)=a+bx+cx^2, \text{show}$

$\int \frac{5x^2 - 4x - 3}{(1 + 2x + 3x^2)^2} dx = \frac{\frac{2+c}{3}+ \frac{2c-5}{3}x+cx^2}{1+2x+3x^2}.$

$\text{Since the choice of }R(x)\text{ is not unique, by comparing the two functions }R(x)$
$\text{corresponding to two different values of }a,\text{ explain how the different choices are related}\text{.}$

Solution:

My work (My latex is below but I didn't get it right)

3{{x}^{2}}+2x+1\overset{\frac{c}{3}}{\overline{\le ft){\begin{align} 
& \text{ }c{{x}^{2}}+\frac{2c-5}{3}x+\frac{2+c}{3} \\
& \underline{-\left( c{{x}^{2}}\text{ }+\text{ }\frac{2c}{3}x+\text{ }\frac{c}{3} \right)} \\
& \text{ }\frac{-5}{3}x\text{ + }\frac{2}{3} \\
\end{align}}\right.}}\text{ so }\frac{R(x)}{Q(x)}=\frac{c}{3}+\frac{1}{3}\left( \frac{-5x+2}{1+2x+3{{x}^{2}}} \right) \\
& \text{But as you can see above, solution has }\frac{R(x)}{Q(x)}=\text{constant + }\frac{-3x-2{{x}^{2}}}{1+2x+3{{x}^{2}}} \\

2. ## Re: My Own Proof of uniqueness of antiderivative

I'm confused. Can't see the LaTeX, don't know what a,b,c are.

Just note that the derivative of a constant function C is zero: $\frac{d}{dx}(C) = 0$ and $\frac{d}{dx} (f(x) + C) = \frac{d}{dx} f(x) = f'(x)$

Therefore $\int f'(x) \, dx$ could equal $f(x)$ or $f(x) + C$ for any constant C. Because for a given anti-derivative, we can shift that function up or down, and it doesn't change the value of $f'(x)$ at that x-coordinate. The same applies to vector-valued functions, where C is an arbitrary constant vector

Edit: I see the solution now, thanks

3. ## Re: My Own Proof of uniqueness of antiderivative

Thank you richard1234, for your response. I understand the ideas mentioned but the question asks for the details for this particular integral above.
Sorry about my post. I fixed it now so everything should be running, fingers crossed.

4. ## Re: My Own Proof of uniqueness of antiderivative

The definition of "derivative" shows that the derivative of f(x)+ C is f'(x) for ant ionstant C. To show the other way, that if f'(x)= g'(x), for all x, then g(x)= f(x)+ C, requires the mean value theorem:

Suppose that, for all x, f'(x)= g'(x). Let F(x)= f(x)- g(x). Then, by the mean value theorem, for any a, $\frac{F(a)- F(0)}{a- 0}= F'(c)$ for some c between 0 and a. That is, $\frac{f(a)- g(a)- (f(0)- g(0))}{a}= f'(c)- g'(c)= 0$. From that, f(a)- g(a)= f(0)- g(0)= C so f(a)= g(a)+ C.

5. ## Re: My Own Proof of uniqueness of antiderivative

Thank you HallsofIvy, but I'm afraid I don't see how I can apply your post to fix my solution to the question above?

Edit: I understand that this question basically relies on the result that $f'(x) = g'(x) \text{ for all } x \implies f(x) = g(x) +C$. However, my troubles aren't with this.

Instead, I'm trying to figure out why my expression gotten from polynomial long division for $\frac{R(x)}Q(x)}}$ is DIFFERENT from the given solution's. They should be the same because it is the integration constant, independent of $x$ that can take on any real number.

I got $\frac{R(x)}{Q(x)}} = \dfrac{c}{3} + \frac{1}{3}(\dfrac{-5x+2}{1 + 2x + 3x^2})$.

Solution has $\frac{R(x)}{Q(x)}} = \text{ constant } + \frac{1}{3}(\dfrac{-3x - 2x^2}{1 + 2x + 3x^2})$