Finding two functions of an indefinite integral

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MooreLikeMike

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The question I'm working on asks to "Find : ∫ (e^(3x-1)e^(2x+2)-3/(4x+1))dx. Find two functions g (x) and h (x) such that g' (x) = (e^(3x-1)e^(2x+2)-3/(4x-1) = h' (x)

I found the antiderivative of the first part of the question: 1/5*e^(5x+1)-3/4*ln(abs(4x+1))+C. I just don't know how I'm supposed to find two functions. I originally thought that I was just supposed to write the function that I found twice, but changing the constant "C" to different numbers. But that doesn't seem right to me. I know that the one function I found satisfies g (x) OR h (x), but not both.
 
What does the integral have to do with the matching derivatives? Is part of the problem statement missing?

"write the function that I found twice, but changing the constant "C" to different numbers. But that doesn't seem right..."

Why do you care how it "seems"? Is it correct? That's kind of the whole point of the "+C", isn't it? Some constant may have disappeared upon differentiation.
 
I know that the one function I found satisfies g (x) OR h (x), but not both OR none. Why can't g(x) = h(x)?


Suppose I say d(x) = x^2 and e(x) = x^2. Did I not list/define two different functions? Sure for this exercise try not to use g(x) = h(x) as it defeats the lesson of the problem, but using the same two functions does not violate anything as being a valid solution.
 
tkhunny said:
What does the integral have to do with the matching derivatives? Is part of the problem statement missing?

"write the function that I found twice, but changing the constant "C" to different numbers. But that doesn't seem right..."

Why do you care how it "seems"? Is it correct? That's kind of the whole point of the "+C", isn't it? Some constant may have disappeared upon differentiation.
Click to expand...
Nothing of the problem statement is missing. I copied it exactly. I guess the reason why I thought it didn't "seem" right was because when I saw that I had to find two functions that would have derivatives that equal each other, I was thinking that they had to be two completely functions. But what you're saying does make sense to me since the whole point of "+C" is that the constant can be any real number, which in turn, would make the functions different if they had different constants.
Thank you!
 
Jomo said:
I know that the one function I found satisfies g (x) OR h (x), but not both OR none. Why can't g(x) = h(x)?


Suppose I say d(x) = x^2 and e(x) = x^2. Did I not list/define two different functions? Sure for this exercise try not to use g(x) = h(x) as it defeats the lesson of the problem, but using the same two functions does not violate anything as being a valid solution.
Click to expand...
You're right, although they're the same, you listed two different functions, which is the lesson of the problem. I guess I just assumed that the functions had to vary in a way
 
No, the lesson of the problem is that two functions that differ by a constant still have the same derivative.

Suppose that integral you listed equaled F(x). Then one answer would be g(x) = F(x) + 7 and h(x) = F(x) + 2/3
 
Jomo said:
No, the lesson of the problem is that two functions that differ by a constant still have the same derivative.

Suppose that integral you listed equaled F(x). Then one answer would be g(x) = F(x) + 7 and h(x) = F(x) + 2/3
Click to expand...
I'm sorry that's what I meant to say. You're saying if two functions have the same derivative "6x^2", then d(x) = 2x^3 +9 and h(x) = 2x^3 + 2 would a valid solution. Is that correct?
 
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