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Help me? please How to solve this integral? $$\int\frac{1+x^2}{1+x^4}\,dx$$

Erka
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    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? – Karolis Juodelė Dec 07 '13 at 11:21
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    Hi @Erka. Welcome! You might find this helpful. – Shaun Dec 07 '13 at 11:22
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    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. – Did Dec 07 '13 at 12:03

2 Answers2

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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

robjohn
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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

lab bhattacharjee
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