- #1
Natasha1
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Am I right in saying that the integral of 1/(x-6)^2 dx
= (x-6)^-1 / -1
= - 1 / (x-6)
= (x-6)^-1 / -1
= - 1 / (x-6)
Simplifying the Integral of 1/(x-6)^2 and Checking for Accuracy
- Thread starter Natasha1
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In summary: When integrating a function, you don't know which of these infinite posible values to C is the one, so you add an integration constant to make sure that the function will give the same answer regardless of which value of C you choose.
- #2
siddharth
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Add the Integration constant, and you are correct.
- #3
Natasha1
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siddharth said:
I will thanks very much :-)... oh could someone explain exactly what this integral constant is? And why is it needed?
- #4
TD
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When we write [itex]\int x dx[/itex] we're looking for the function(s) which give x again if we take the derivative. Now, you know this integral is x²/2 since (x²/2)' = x but when you add a constant, it's still valid: (x²/2 + c)' = x. This is because the derivative of a constant is zero.
- #5
Natasha1
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TD said:
thanks TD for your explanation
- #6
TD
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You're welcome 
- #7
rea
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If you have a function
f(x) = x^2 there are other functions of the same shape that difer only in the position respect "y"...
if f(x)=x; is like if you have f(x)=x+0=x+C, the +C let yo translate up or down the graph of the function.
Is necesary when integrating a function, because you don't know which of this infinite posible values to C is the one, thus, instead of write a infinite set of functions that only difer in the a real number "you select" one specific function of those many posibles that represent such infinite funtions and call it F(x)+C, thus ;), you have one function that can represent any function with the same shape and that only difer in the y position instead of have a infinite number of functions to analise.
f(x) = x^2 there are other functions of the same shape that difer only in the position respect "y"...
if f(x)=x; is like if you have f(x)=x+0=x+C, the +C let yo translate up or down the graph of the function.
Is necesary when integrating a function, because you don't know which of this infinite posible values to C is the one, thus, instead of write a infinite set of functions that only difer in the a real number "you select" one specific function of those many posibles that represent such infinite funtions and call it F(x)+C, thus ;), you have one function that can represent any function with the same shape and that only difer in the y position instead of have a infinite number of functions to analise.
FAQ: Simplifying the Integral of 1/(x-6)^2 and Checking for Accuracy
What is the process for simplifying the integral of 1/(x-6)^2?
The first step is to use the substitution u = x-6, which will transform the integral into 1/u^2. Then, use the power rule for integration to get -1/u + C. Finally, substitute back in for u and simplify the expression.
Can the integral of 1/(x-6)^2 be simplified further?
No, the integral cannot be simplified any further. The final expression should be in the form -1/(x-6) + C.
How do you check the accuracy of the simplified integral?
To check the accuracy, you can differentiate the simplified integral and see if it matches the original integrand. If it does, then the simplified integral is correct.
Are there any common mistakes to watch out for when simplifying this integral?
One common mistake is forgetting to substitute back in for u after using the substitution method. Another mistake is not applying the power rule for integration correctly.
Is there a specific domain in which this simplified integral is valid?
Yes, the simplified integral is only valid for x ≠ 6, as the original integrand is undefined at x = 6. So the domain of the simplified integral is (-∞, 6) ∪ (6, ∞).