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 ofand 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.
Solution:
My work (My latex is below but I didn't get it right)
{\begin{align})
} \\ <br />
& \text{ }\frac{-5}{3}x\text{ + }\frac{2}{3} \\ <br />
\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) \\ <br />
& \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}}} \\ <br />
)
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:and
Thereforecould equal
or
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
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
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.
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,for some c between 0 and a. That is,
. From that, f(a)- g(a)= f(0)- g(0)= C so f(a)= g(a)+ C.
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. However, my troubles aren't with this.
Instead, I'm trying to figure out why my expression gotten from polynomial long division foris DIFFERENT from the given solution's. They should be the same because it is the integration constant, independent of
that can take on any real number.
I got.
Solution has}{Q(x)}} = \text{ constant } + \frac{1}{3}(\dfrac{-3x - 2x^2}{1 + 2x + 3x^2}))
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