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Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

$ \displaystyle g(x) = \int^x_1 \ln (1 + t^2) \,dt $

$g^{\prime}(x)=\ln \left(1+x^{2}\right)$

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All right, so let's move on to the next question. We're still using the same concept to fundamental form of calculus. We have a different integral, so it is going to be the G of X. A function of X is equal to the integration from 12 x natural. Log off one plus T squared DT. So I want you to listen to me carefully, OK, What this means this expression means is that it is the anti derivative of the natural log of one plus something squared, and then it's going to end up being a function off X. That's how I want you to interpret this expression right there. So again, it's an anti derivative off right here. That's a function of X. So if I take the derivative with respect to X, the anti the derivative off, the anti derivative are going to undo each other, so you will just get what's inside and remember, it's a function of X. So instead of a T square, you will see an X squared, and this is how you do this problem

University of California, Berkeley