# How to change variables in integration of y=(x^2-1)/(x^4+1)? indication x+(1/x)=u

*print*Print*list*Cite

Determine `int (x^2-1)/(x^4+1) dx` :

Rewrite as `int (((x^2-1)dx)/x^2)(1/(x^2+1/x^2))`

Let `u=x+1/x`

Then `du=(1-1/x^2)dx=(x^2-1)/x^2 dx`

Also `u^2=x^2+1/x^2+2` so `1/(x^2+1/x^2+2-2)=1/(u^2-2)`

So `int (x^2-1)/(x^4+1) dx=int (du)/(u^2-2)` '

`int (du)/(u^2-2)=int (du)/((u+sqrt(2))(u-sqrt(2)))`

We can use partial fractions:

`A/(u+sqrt(2))+B/(u-sqrt(2))=1/(u^2-2)`

`==> Au-sqrt(2)A+Bu+sqrt(2)B=1`

`==> A+B=0,-sqrt(2)A+sqrt(2)B=1`

So `B=1/(2sqrt(2)),A=-1/(2sqrt(2))`

Then `int (du)/(u^2-1)=-1/(2sqrt(2)) int (du)/(u+sqrt(2))+1/(2sqrt(2)) int (du)/(u-sqrt(2))`

`=-1/(2sqrt(2))ln|u+sqrt(2)|+1/(2sqrt(2))ln|u-sqrt(2)|+C`

Substituting for u we get:

`=-1/(2sqrt(2))ln|x+1/x+sqrt(2)|+1/(2sqrt(2))ln|x+1/x-sqrt(2)|+C`

`=1/(2sqrt(2))ln|(x+1/x-sqrt(2))/(x+1/x+sqrt(2))|+C`