# AP Calculus BC

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Hi! Thank you very much for your help---
I'm not sure what the answer to this is; how do I solve?

Find antiderivative of
(1/(x^2))[sec(1/x)][tan(1/x)]dx

I did integration by parts and got to
(1/(x^2))[sec(1/x)] + 2*[antiderivative of (1/(x^3))(sec(1/x))dx]

• AP Calculus BC -

Integration by parts is the same as any other tool. It's just a tool. You can go around in circles with it... unless you know where you're going.

For this particular problem, I propose to use another tool, substitution.

Did you notice there is the factor (1/x²) at the beginning? What would ∫(1/x²)dx give? ∫-d(1/x).

So the integral becomes:
I=∫(1/(x^2))[sec(1/x)][tan(1/x)]dx
=∫[sec(1/x)][tan(1/x)]d(1/x)
=∫sec(y)tan(y)dy
= ... +C

Do remember, however, if and when you have to evaluate a definite integral, the limits have to correspond to the integration variable, which in this case is (1/x).

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