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# Find $f$.$f"(t) = x^{-2}$, $\quad x > 0$, $\quad f(1) = 0$, $\quad f(2) = 0$

## $$f(x)=-\ln |x|+\ln 2 x-\ln 2$$

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Okay, The first thing we know we can do is we can take the first integral, which means our new exponents becomes to the negative one. We're dividing by the new exponents negative one, which gives us one over negative x pussy. So now we know that we can plug end to get our constant C and deep. Remember, the negative natural law of one is just zero because the natural of one is just zero, therefore got cancels. Therefore, we end up with d equals negative c. This is important because now we're doing half of twos. Negative, naturally of two plus two C plus d, which gives zero is negative. Natural log of two pussy which gives us the natural log of two equals c right, because we're transferring this over to the left hand side. Therefore, changing a symbol which gives our final equation as affects is negative not for a lot of acts, plus Ellen of two x minus natural order, too

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