First, let us consider the following Integrals :
#I_1=int(x^2+1)/(x^4+1)dx, and, I_2=int(x^2-1)/(x^4+1)dx.#
Now, #I_1=int{x^2(1+1/x^2)}/{x^2(x^2+1/x^2)}dx,#
#:. I_1=int(1+1/x^2)/(x^2+1/x^2)dx.#
Subst., #x-1/x=u," so that, "(1+1/x^2)dx=du," and, also, "#
# (x^2+1/x^2)=(x-1/x)^2+2=u^2+2.#
#:. I_1=int1/(u^2+2)du=1/sqrt2arc tan(u/sqrt2).#
As, #u=x-1/x=(x^2-1)/x,# we have,
# I_1=1/sqrt2arc tan((x^2-1)/(sqrt2*x)).....................(1).#
Next, #I_2=int(x^2-1)/(x^4+1)dx=int{x^2(1-1/x^2)}/{x^2(x^2+1/x^2)}dx,#
#:. I_2=int(x^2-1/x^2)/(x^2+1/x^2)dx.#
Take #x+1/x=v," so that, "(1-1/x^2)dx=dv," and, further, "#
#x^2+1/x^2=(x+1/x)^2-2=v^2-2.#
#:. I_2=int1/{v^2-(sqrt2)^2}dv=1/(2sqrt2)ln|(v-sqrt2)/(v+sqrt2)|.#
Replacing #v, by, x+1/x=(x^2+1)/x,# we get,
#I_2=1/(2sqrt2)ln|(x^2-sqrt2*x+1)/(x^2-sqrt2*x+1)|......(2).#
Finally, observe that, the Reqd. Integral #I,# is,
#I=int1/(x^4+1)dx=1/2int{(x^2+1)-(x^2-1)}/(x^4+1)dx,#
#rArr I=1/2*I_1-1/2*I_2,#
#=1/(2sqrt2)arc tan((x^2-1)/(sqrt2*x))-1/(4sqrt2)ln|(x^2-sqrt2*x+1)/(x^2-sqrt2*x+1)|+C.#
N.B. : #J=intx^2/(x^4+1)dx=1/2*I_1+1/2*I_2.#
Enjoy Maths.!
