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While studying for an exam I came across a trig substitution integration problem without a square root and I'm not sure how to approach it.

$$\int \frac{1}{1+3x^2} \, dx $$

Any ideas?

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  • $\begingroup$ the first thing i got to my mind is use $x = tan(\theta)$ and $1 = 3-2$ $\endgroup$ – richitesenpai Apr 13 '15 at 22:16
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This looks like a straightforward $\arctan$ integral. Try writing it in the form $$ \int \frac{f'(x) } {1+[f(x)]^2} \mathrm{d} x. $$

It should be clear to you, comparing the denominators, that $f(x)=\sqrt{3}x$.

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