Simplifying the Integral of 1/(x-6)^2 and Checking for Accuracy

[Natasha1]

Am I right in saying that the integral of 1/(x-6)^2 dx

= (x-6)^-1 / -1

= - 1 / (x-6)

 

[siddharth]

Add the Integration constant, and you are correct.

 

[Natasha1]

Add the Integration constant, and you are correct.

I will thanks very much :-)... oh could someone explain exactly what this integral constant is? And why is it needed?

 

[TD]

When we write \int x dx 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.

 

[Natasha1]

When we write \int x dx 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.

thanks TD for your explanation

 

[TD]

You're welcome

 

[rea]

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.