Let, I=intx^23/(x^32+1)dx.
d/dx(x^24)=24x^23 suggests that the substitution x^24=y may work.
x^24=y rArr 24x^23dx=dy.
:. I=1/24int1/{(y^(1/24))^32+1}dy=1/24int1/(y^(4/3)+1)dy.
To get rid of cube root, we let,
y^(1/3)=t rArr 1/3*y^(-2/3)dy=dt rArr dy=3y^(2/3)dt=3t^2dt.
:. I=3/24intt^2/(t^4+1)dt=1/8intt^2/(t^4+1)dt,
=1/4int(2t^2)/(t^4+1)dt=1/4int{(t^2+1)+(t^2-1)}/(t^4+1)dt.
:. 4I=int(t^2+1)/(t^4+1)dt+int(t^2-1)/(t^4+1)dt=I_1+I_2, say.
I_1=int(t^2+1)/(t^4+1)dt=int{t^2(1+1/t^2)}/{t^2(t^2+1/t^2)dt.
:. I_1=int(t^2+1/t^2)/(t^2+1/t^2)dt=int{d/dt(t-1/t)}/{(t-1/t)^2+2}dt,
=int(du)/(u^2+2)............[u=t-1/t],
=1/sqrt2arc u/sqrt2,
=1/sqrt2arc ((t^2-1)/(sqrt2t)),
=1/sqrt2arc ((y^(2/3)-1)/(sqrt2*u^(1/3))),
=1/sqrt2arc ((x^16-1)/(sqrt2*x^8)).
Further, I_2=int(t^2-1)/(t^4+1)dt
=int{t^2(1-1/t^2)}/{t^2(t^2+1/t^2)}dt,
=int(1-1/t^2)/(t^2+1/t^2)dt,
=int{d/dt(t+1/t)}/{(t+1/t)^2-2}dt,
=int(dv)/(v^2-2)dv.............[v=t+1/t],
=1/(2sqrt2)ln|(v-sqrt2)/(v+sqrt2)|,
=1/(2sqrt2)ln|(t^2-sqrt2t+1)/(t^2+sqrt2t+1)|,
=1/(2sqrt2)ln|(y^(2/3)-sqrt2*y^(1/3)+1)/(y^(2/3)+sqrt2*y^(1/3)+1)|,
=1/(2sqrt2)ln|(x^16-sqrt2*x^8+1)/(x^16+sqrt2*x^8+1)|.
Finally, we get,
I=1/(4sqrt2)arc ((x^16-1)/(sqrt2*x^8))
+1/(8sqrt2)ln|(x^16-sqrt2*x^8+1)/(x^16+sqrt2*x^8+1)|+C.
Enjoy Maths.!
