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.I1=∫x2+1x4+1dx,and,I2=∫x2−1x4+1dx.
Now, I_1=int{x^2(1+1/x^2)}/{x^2(x^2+1/x^2)}dx,I1=∫x2(1+1x2)x2(x2+1x2)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.!
