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$$\int_2^3 \frac{1+x^2}{1+x^4}~ dx$$

I am trying this question by the method of separating. $[1/(1+x^4) + x^2/(1+x^4)]$ But how to proceed further because if I take $x^2=t$ then the above integrand will be $[1/(2\sqrt {t/(1+t²)}+(1/2)(\sqrt{t/(1+t^2)}]$. But how to proceed further?

Blue
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    $1+x^4=1+2x^2+x^4-2x^2=(1+x^2)^2-2x^2=(1+\sqrt{2}x+x^2)(1-\sqrt{2}x+x^2)$. You can write the integrand as $\frac{Ax+B}{1+\sqrt{2}x+x^2}+\frac{Cx+D}{1-\sqrt{2}x+x^2}$ for some constants $A,B,C,D$. Do you know how to integrate these simpler fractions $\frac{ax+b}{1\pm\sqrt{2}x+x^2}$ ? – NDB Jul 17 '23 at 18:26
  • Substituting $y=\sqrt{\frac{1+x}{1-x}}$ produces an integral that is evaluated here.$$\int_2^3 \frac{1+x^2}{1+x^4} , dx = \int_\tfrac35^\tfrac45 \frac{dy}{\left(1+y^2\right) \sqrt{1-y^2}}$$ – user170231 Apr 25 '24 at 21:46

3 Answers3

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$$\int \frac{1+x^2}{1+x^4}dx$$ $$=\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx$$ $$=\int \frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2-2}dx$$ $$=\int \frac{dy}{y^2+2}$$ where $y=x-\frac{1}{x}$

The last integrand is pretty easy to solve. I hope you can take it from here.

Blue Cat Blues
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$$x^4+1=x^4+2x^2+1-2x^2=$$ $$=(x^2+1)^2-2x^2=(x^2-\sqrt2 x+1)(x^2+\sqrt2x+1)$$ using partial fractions $$\frac{x^2+1}{x^4+1}=\frac{Ax+B}{x^2-\sqrt2 x+1}+\frac{Cx+D}{x^2+\sqrt2 x+1}$$ then

$A=C=0$ and $B=D=\frac{1}{2}$

$$\int \frac{x^2+1}{x^4+1}=\int \frac{1}{2}(\frac{1}{x^2-\sqrt2 x+1}+\frac{1}{x^2+\sqrt2 x+1})=$$ $$=\int \frac{1}{2}(\frac{1}{(x-\frac{1}{\sqrt2})^2 + \frac{1}{2}}+\frac{1}{(x+\frac{1}{\sqrt2})^2 + \frac{1}{2}})=$$ $$=\int \frac{1}{(\sqrt2x-1)^2+1} + \frac{1}{(\sqrt2x+1)^2+1}=$$ $$=\int \frac{1}{\sqrt2}(\frac{\ du}{1+u^2}+\frac{\ dv}{1+v^2})=$$ $$= \frac{1}{\sqrt2}(\arctan u +\arctan v)=$$ $$=\frac{1}{\sqrt2}(\arctan (\sqrt2x-1)+\arctan (\sqrt2x+1)) + c$$

$u=\sqrt2x-1,\ du=\sqrt2\ dx$

$v=\sqrt2x+1,\ dv=\sqrt2\ dx$

Lion Heart
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Here is another interesting non standard way that works for the following types: $$\int \frac{1}{x^4+ax^2+1},\int \frac{x^2}{x^4+ax^2+1},\int \frac{x^2\pm1}{x^4+ax^2+1}$$

\begin{gather} I = \int \frac{1+x^2}{1+x^4}\\ = \int \frac{1}{1+x^4} + \int\frac{x^2}{1+x^4}\\ = \int \frac{\frac{1}{x^2}}{\frac{1}{x^2}+x^2} + \int\frac{1}{\frac{1}{x^2}+x^2}\\ =\int \frac{\frac{1}{2}((1+\frac{1}{x^2})-(1-\frac{1}{x^2}))}{\frac{1}{x^2}+x^2}+\int \frac{\frac{1}{2}((1+\frac{1}{x^2})+(1-\frac{1}{x^2}))}{\frac{1}{x^2}+x^2}\\ = \frac{1}{2}(\int \frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+2}-\int \frac{1-\frac{1}{x^2}}{(x+\frac{1}{x})^2-2}+\int \frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+2}+\int \frac{1-\frac{1}{x^2}}{(x+\frac{1}{x})^2-2})\\ =\int \frac{1+\frac{1}{x^2}}{(x-\frac{1}{x})^2+2}\\ \text{Let} \ x-\frac{1}{x}= t \ ; \ dt=1+\frac{1}{x^2}\\ =\int \frac{dt}{(t)^2+2} = \frac{1}{\sqrt{2}}(arctan(\frac{t}{\sqrt2}))\\ \fbox{I = $\frac{1}{\sqrt{2}}$(arctan($\frac{x-\frac{1}{x}}{\sqrt 2}$))}\\ \text{Here it was lengthy but when $a\neq 0$ this method is very useful} \end{gather}

XZCY
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