# How do you use the second fundamental theorem of Calculus to find the derivative of given #int [(ln(t)^(2))/t]dt# from #[3,x]#?

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Instead of actually finding the function and then differentiating, one could reason as follows:

The Second Fundamental Theorem of Calculus says that

NOTE

The First Fundamental Theorem of Calculus says this directly. It says:

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To find the derivative of the integral ( \int_{3}^{x} \frac{\ln(t)^2}{t} , dt ), you use the second fundamental theorem of calculus, which states that if ( f(x) ) is continuous on ([a, b]) and ( F(x) ) is an antiderivative of ( f(x) ) on ([a, b]), then:

[ \frac{d}{dx} \left( \int_{a}^{x} f(t) , dt \right) = f(x) ]

First, find an antiderivative ( F(x) ) of ( \frac{\ln(t)^2}{t} ). Let ( u = \ln(t) ), then ( du = \frac{1}{t} dt ). So, ( \int u^2 , du = \frac{u^3}{3} + C = \frac{\ln(t)^3}{3} + C ).

Therefore, ( F(x) = \frac{\ln(x)^3}{3} + C ).

Now, apply the second fundamental theorem of calculus:

[ \frac{d}{dx} \left( \int_{3}^{x} \frac{\ln(t)^2}{t} , dt \right) = \frac{d}{dx} \left( F(x) - F(3) \right) = \frac{d}{dx} \left( \frac{\ln(x)^3}{3} - \frac{\ln(3)^3}{3} \right) ]

[ = \frac{\frac{3\ln(x)^2}{x}}{3} = \frac{\ln(x)^2}{x} ]

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