Indefinite integral (03)

[Grigorios Kostakos]

Evaluate \[\displaystyle\int{\frac{\sqrt{1+x^4}}{x^4-1}\,dx}\,.\]

Hint

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Use Indefinite integral (02)

[jacks]

\[\int\frac{\sqrt{x^4+1}}{x^4-1}dx=\int\frac{x^4+1}{\left(x^4-1\right).\sqrt{x^4+1}}dx\]
\[=\int\frac{\left(x^2+1\right)^2-2x^2}{\left(x^4-1\right).\sqrt{x^4+1}}dx=\int\frac{x^2+1}{\left(x^2-1\right).\sqrt{x^4+1}}dx-\int\frac{2x^2}{\left(x^2-1\right).\sqrt{x^4+1}}dx\]
\[=\int\frac{x^2+1}{\left(x^2-1\right).\sqrt{x^4+1}}dx-\frac{1}{2}.\int\left(\frac{x^2+1}{x^2-1}-\frac{x^2-1}{x^2+1}\right).\frac{1}{\sqrt{x^4+1}}dx\]
\[=\frac{1}{2}\int\frac{x^2+1}{\left(x^2-1\right).\sqrt{x^4+1}}dx+\frac{1}{2}\int\frac{x^2-1}{\left(x^2+1\right).\sqrt{x^4+1}}dx\]