Question

Evaluate:  ${\displaystyle \int \frac{x(1+x^2)}{1+x^4}dx}$

Answer 1

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

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

Let 1+x² t
      2x dx = dt
      ${\displaystyle xdx=\frac{1}{2}dt}$

     ${\displaystyle I=\frac{1}{2}\int \frac{t\times dt}{t^2-2(t-1)}}$

${\displaystyle I=\frac{1}{2}\int \frac{t dt}{t^2-2t+2}}$

${\displaystyle I=\frac{1}{2\times 2}\int \frac{2t dt}{t^2-2t+2}}$

${\displaystyle I=\frac{1}{4}\int \frac{(2t-2)+2dt}{t^2-2t+2}}$

${\displaystyle I=\frac{1}{4}(\frac{\int (2t-2)+2dt}{t^2-2t+2}dt+2\int \frac{dt}{t^2-2t+2})}$

${\displaystyle I=\frac{1}{4}In|t^2-2t+2|+\frac{2}{4} \int \frac{dt}{{(t-1)}^2}+C_1}$

${\displaystyle =\frac{1}{4}In|t^2-2t+2|+\frac{1}{2} {\tan }^{-1}(t-1)+c_1}$

${\displaystyle \therefore 1+X^2=t}$

${\displaystyle =\frac{1}{4}In|(1+x^2)-2(1+x^2)+2|+\frac{1}{2}{\tan }^{-1}(1+x^2-1)+C}$

=${\displaystyle \frac{1}{4}In|1+x^4+2x^2-2(1+x^2)|+\frac{1}{2}{\tan }^{-1}\left(x^2\right)}$

${\displaystyle =\frac{1}{4}In|X^4+1|+\frac{1}{2}{\tan }^{-1}\left(x^2\right)+C}$