Prove Integral f&g Defined in R: Find Derivative

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SUMMARY

The discussion centers on proving the definitions and derivatives of the functions F(x) and G(x) based on the functions f and g. The function f is defined as f(x) = 1/((ln(x+1))^2 + 1) for x > -1 and f(-1) = 0. The integral F(x) = ∫(0 to x^2 + 2x) f(t) dt is confirmed to be defined for all x in R, with its derivative calculated as F'(x) = (2x + 2)/((ln(x^2 + 2x + 1))^2 + 1). The function g is defined similarly, and it is established that G(x) = ∫(0 to x^2 + 2x) g(t) dt is also defined for all x > -1, sharing the same derivative as F(x).

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Homework Statement


f is a function defined as: for all x>-1 f(x) = 1/((ln(x+1))^2 + 1) and for x=-1 f(x)=0.
a) Prove that [tex]F(x) = \int^{x^2 + 2x}_{0} f(t)dt[/tex] is defined in R and has a derivative there. Find the derivative.

b) g is defined as: for all x>-1 g(x) = f(x) and for x=-1 g(x) = -1.
Is [tex]G(x) = \int^{x^2 + 2x}_{0} g(t)dt[/tex] defined in R? Does it have a derivative there?

Homework Equations





The Attempt at a Solution


a)first I proved that f is continues for all x>=-1 Then since x^2+2x>0 for all x in R F(x) is defined and it's derivative is easy to get with the chain rule. I got:
F'(x) = (2x+2)/((ln(x^2 + 2x +1))^2 + 1)

b)Since g differs from f in only one point F(x) = G(x) and so G is defined and has a derivative for all x in R.

Is that right? I'm mostly worried about (b).
Thanks.
 
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Looks good to me :) Other than the small error than G is defined and has a derivative for all x > -1, not all real values.
 
In (2), it helps to notice, as you said for (1), that x2+ 2x is never negative: the interval of integration, 0 to x2+ 2x, does not contain -1 and so what happens there is irrelevant.
 

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