Integration

**roshanhero** · March 31st 2010, 10:12 PM

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

**Anonymous1** · March 31st 2010, 10:16 PM

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

oops..

**Anonymous1** · March 31st 2010, 10:24 PM

Another hint:

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

**drumist** · March 31st 2010, 10:42 PM

Originally Posted by Anonymous1

Hint:
$\int \frac{1}{1+x^4} = tan^{-1}(x^2)$

This is NOT correct.

But we can observe that:

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

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

**mr fantastic** · March 31st 2010, 11:06 PM

Originally Posted by roshanhero

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

Use the substitution $u = x^2$ .

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

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