• anonymous
Question 10 18.01 Practice Questions for Exam 2... What is going on in the solution? I follow everything to do with putting all the y on the left of the equation and all the x on the right of the equation... but then boom they use tan instead of integrating??? What is happening
MIT 18.01 Single Variable Calculus (OCW)
  • Stacey Warren - Expert
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  • chestercat
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  • phi
the integral evalutes to \[ \int \frac{dx}{1+x^2}= \tan^{-1} x +C\] See the notes for lecture 5 where they derive \[ \frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}\] and if we take the integral of both sides, we get the formula they used.

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