Help me? please How to solve this integral? $$\int\frac{1+x^2}{1+x^4}\,dx$$
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3$\begingroup$ There is a very strictly defined list of basic steps to make, to integrate a rational function. Are you aware of it? Which one are you having problems with? $\endgroup$– Karolis JuodelėCommented Dec 7, 2013 at 11:21
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2$\begingroup$ Hi @Erka. Welcome! You might find this helpful. $\endgroup$– ShaunCommented Dec 7, 2013 at 11:22
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1$\begingroup$ This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. $\endgroup$– DidCommented Dec 7, 2013 at 12:03
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2 Answers
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$$
\begin{align}
\int\frac{1+x^2}{1+x^4}\,\mathrm{d}x
&=\frac12\int\frac{\mathrm{d}x}{1-\sqrt2x+x^2}+\frac12\int\frac{\mathrm{d}x}{1+\sqrt2x+x^2}\tag{1}\\
&=\int\frac{\mathrm{d}x}{\left(\sqrt2x-1\right)^2+1}+\int\frac{\mathrm{d}x}{\left(\sqrt2x+1\right)^2+1}\tag{2}\\
&=\frac1{\sqrt2}\left(\tan^{-1}(\sqrt2x-1)+\tan^{-1}(\sqrt2x+1)\right)+C\tag{3}\\
&=\frac1{\sqrt2}\tan^{-1}\left(\frac{\sqrt2x}{1-x^2}\right)+C\tag{4}
\end{align}
$$
Explanation:
$(1)$: partial fractions
$(2)$: complete square
$(3)$: arctan integral
$(4)$: tan of a sum formula
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HINT:
$$\displaystyle \int\frac{1+x^2}{1+x^4}dx=\int\frac{\frac1{x^2}+1}{x^2+\frac1{x^2}}dx$$
As $\displaystyle\int\left(\frac1{x^2}+1\right)dx=x-\frac1x$
write $\displaystyle x^2+\frac1{x^2}=\left(x-\frac1x\right)^2+2 $ and set $\displaystyle x-\frac1x=u$
Then use Trigonometric substitution
answered Dec 7, 2013 at 13:39
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$\begingroup$ @Erka, May I request you to establish the equivalence of the two answers? $\endgroup$ Commented Dec 7, 2013 at 15:50