Suppose the function f is defined by f(x)= integral from sqrt x to 1 of (2t - 1)/(t + 2) dt for all real numbers x > or = 0?
a) evaluate f(1).
b) evaluate f'(1).
c) find an equatino for the tangent line to the graph of f at the tpoint where x=1.
d) on what intervals is the function f increasing? justify your answer

F(x)= ∫[√x, 1] (2t - 1)/(t + 2) dt for x ≥ 0 

Let f(x) = ∫(2t - 1)/(t + 2) dt = ∫[2 + -5/(t + 2)] dt = 2t - 5Ln(t + 2) + constant 

F(x) = f(1) - f(√x) = 2 – 5Ln(3) - (2√x - 4Ln(√x + 2) 

F(1) = 2 - 5Ln(3) - (2√1 - 5Ln(√1 + 2) = 2 - 5Ln(3) - (2 - 5Ln(3)) = 0

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The answer is lol i dont know