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How do I evaluate $$\int_{0}^{1} \frac{x^{2} + 1}{x^{4} + 1 } \ dx$$

I tried using substitution but I am getting stuck. If there was $x^3$ term in the numerator, then this would have been easy, but this one doesn't.

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up vote 12 down vote accepted


  • Try dividing the numerator and denominator by $x^{2}$.

  • Then you get $\displaystyle \int\frac{1 + \frac{1}{x^2}}{x^{2}+\frac{1}{x^2}} \ dx$.

  • Write $\displaystyle x^{2} +\frac{1}{x^2}$ as $\displaystyle \biggl(x-\frac{1}{x}\biggr)^{2} + 2$

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thanks. Never thought of this – Romeo Jun 8 '12 at 15:41
You don't solve integrals, you evaluate integrals. – Stefan Smith Jun 8 '12 at 15:49

Another way, if you want to sweat harder instead of the elegant suggestion of Chandrasekhar:$$x^4+1=(x^2+\sqrt{2}\,x+1)(x^2-\sqrt{2}\,x+1)\Longrightarrow$$$$ \frac{x^2+1}{x^4+1}=1-\frac{\sqrt 2\,x}{x^2+\sqrt 2\,x+1}+1+\frac{\sqrt 2\,x}{x^2-\sqrt 2\,x+1}$$ so for example$$\int\frac{\sqrt 2\,x}{x^2+\sqrt 2\,x+1}dx=\frac{1}{\sqrt 2}\int\frac{2x+\sqrt 2}{x^2+\sqrt 2\,x+1}dx-\frac{1}{2\sqrt 2}\int\frac{\sqrt 2dx}{(\sqrt 2 x+1)^2+1}=$$$$=\frac{1}{\sqrt 2}\log|x^2+\sqrt 2\,x+1|-\frac{1}{2\sqrt 2}\arctan(\sqrt 2\,x+1)+C$$ and etc.

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@Romeo: He did it practical here. – Babak S. Jun 8 '12 at 16:59
+1 for both this answer and the cleverer one. This one is how you do it if you just follow what all the books say. – Michael Hardy Jun 9 '12 at 3:16

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