Integral of #(x^23)/(x^32+1)#?

1 Answer
Mar 18, 2018

Answer:

#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#.

Explanation:

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.!