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Solve: $$\int\frac{1}{4x^2+4x+2}dx$$

I think it may related to $\tan^{-1}x$. Any hint?

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closed as off-topic by user223391, user91500, Claude Leibovici, Watson, Qwerty Sep 25 '16 at 13:25

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  • $\begingroup$ Is it $4x^4+4x^2+1$ or $4x^4+4x+1$? $\endgroup$ – Jacky Chong Sep 25 '16 at 6:10
  • $\begingroup$ It is $4x^4+4x+2$ $\endgroup$ – Han Tang Sep 25 '16 at 6:11
  • $\begingroup$ Wolframalpha doesn't seem to give a nice solution. $\endgroup$ – Jacky Chong Sep 25 '16 at 6:12
  • $\begingroup$ What did you try? $\endgroup$ – Carl Schildkraut Sep 25 '16 at 6:13
  • $\begingroup$ If the roots are ugly, then the solution must be ugly as well. There is no going around that. $\endgroup$ – Ivan Neretin Sep 25 '16 at 6:14
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Hint: The denominator turns into $$(2x+1)^2+1$$

so you can make the substitution $$y = 2x+1$$ to turn the integral into

$$\frac{1}{2}\int \frac{1}{y^2+1}\ dy$$

Then try to make a trigonometric substitution.

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