1. ## Integration

$\displaystyle \int \frac{x(1+x^2)}{1+x^4)}$
I tried to do it by substitution but reached nowhere

2. Hint:
$\displaystyle \int \frac{1}{1+x^2} = tan^{-1}(x)$

oops..

3. Another hint:

$\displaystyle \int \frac{x+x^3}{1+x^4} = \int \frac{x}{1+x^4} + \int \frac{x^3}{1+x^4}$

4. Originally Posted by Anonymous1
Hint:
$\displaystyle \int \frac{1}{1+x^4} = tan^{-1}(x^2)$
This is NOT correct.

But we can observe that:

$\displaystyle \frac{d}{dx} \tan^{-1}(x^2) = \frac{2x}{1+(x^2)^2}$

However, if you didn't recognize this, you'd have to do a bit more work with partial fractions, etc.

5. Originally Posted by roshanhero
$\displaystyle \int \frac{x(1+x^2)}{1+x^4)}$
I tried to do it by substitution but reached nowhere
Use the substitution $\displaystyle u = x^2$.

If you need more help, please show your work and say where you get stuck.