Question

$\int \frac{1+x^2}{1+x^4}dx$

Answer 1

$I=\int \frac{1+x^2}{1+x^4}dx$

$I=\int \frac{\frac{1}{x^2}+\frac{x^2}{x^4}}{\frac{1}{x^2}+\frac{x^4}{x^2}}dx$

$I=\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx$

$I=\int \frac{1+\frac{1}{x^2}}{(x^2+\frac{1}{x^2}-2+2)}dx=\int \frac{1+\frac{1}{x^2}}{(x-\frac{1}{x}{)}^2+2}$

Put $x-\frac{1}{x}=t⟹(1+\frac{1}{x^2})dx=dt$

Therefore,

$I=\int \frac{dt}{t^2+2}=\int \frac{dt}{t^2+(\sqrt{2}{)}^2}=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+C$

$=\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\frac{x^2-1}{\sqrt{2}x}\right)+C$