# What is integral improper in x=0, x=+infintty, y=dx/x^4+1?

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int^(oo)_0 (dx)/(x^4+1)

We can use partial fractions with a trick.

`1/(x^4+1)=1/(x^4+2x^2+1-2x^2)=1/((x^2+1)^2-2x^2)=`

`1/(((x^2+1)-sqrt(2)x)((x^2+1)+sqrt(2)x))`

Now we can use partial fractions to turn

`1/((x^2-sqrt(2)x+1)(x^2+sqrt(2)x+1))= A/(x^2-sqrt(2)x+1)+(Bx)/(x^2-sqrt(2)x+1)+`

`C/(x^2+sqrt(2)x+1)+(Dx)/(x^2+sqrt(2)x+1) `

I assume you can do partial fractions and get

`A=-sqrt(2)/4` , `C=sqrt(2)/4` and `B=D=1/2`

We also have to complete the square on the bottom

`x^2-sqrt(2)x+1/2+1/2=(x-sqrt(2)/2)^2+1/2=1/2(2(x-sqrt(2)/2)^2+1)` and

`x^2+sqrt(2)x+1/2+1/2=(x+sqrt(2)/2)^2+1/2=1/2(2(x+sqrt(2)/2)^2+1)`

Then we need to integrate

`-sqrt(2)/4*2/(2(x-sqrt(2)/2)^2+1), sqrt(2)/4*2/(2(x+sqrt(2)/2)^2+1)` and the two logrithmic integrals.

When you do this we get

We get `int 1/(x^4+1)=sqrt(2)/4(arctan(sqrt(2)x+1)+arctan(sqrt(2)x-1)+1/2ln((x^2+sqrt(2)x+1)/(x^2-sqrt(2)x+1)))`

Evaluating this a x=0 we get 0

`lim_(x->oo) arctan(sqrt(2)x+1)=pi/2`

`lim_(x->oo) arctan(sqrt(2)x-1)=pi/2`

And `lim_(x->oo) ln((x^2+sqrt(2)x+1)/(x^2-sqrt(2)x+1))=0`

So

`int^(oo)_0 1/(x^4+1) = sqrt(2)/4(pi/2+pi/2+1/8*0) = (sqrt(2)pi)/4`