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$ dy = \frac{1}{1+x^{2}} dx $

How would I integrate this ? Can I use a u substitution?

let $ u = 1+x^{2} dx $

$ du = 2x dx $

am not sure how to go from here?

Any help appreciated thanks

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    $\begingroup$ $\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}$ $\endgroup$
    – E.H.E
    Jun 14, 2015 at 11:44

2 Answers 2

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The integral of $\frac{1}{1+x^2}$ is equal to $\tan^{-1}(x)$, so you can evaluate your expression to give $y = \tan^{-1}(x) + c$

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The solution is $y=\arctan x+c$ using the standard integral

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