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$$\int \frac{dx}{1+x^2} $$ I'm having a hard time knowing what to do when it's just dx on top

A.Bree
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2 Answers2

6

If this was your confusion: the integral is precisely by definition equal to $$\int \frac{1}{1+x^2} \ \mathrm{d}x$$ which is a very well-known integral.

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$$\int \frac{1}{1+x^2} \space dx = \tan^{-1}(x) + C$$

Art
  • 629