Compute integral x^3/(x^8+1)dx from 0 to 1 ?

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Answer 1 · 2018-06-05T09:55:30

#intx^3/(x^8+1)dx=pi/16#

#I=intx^3/(x^8+1)dx=int1/((color(blue)(x^4))^2+1)*color(brown)(x^3dx)#

Subst. #color(blue)(x^4=u)=>4x^3dx=du=>color(brown)(x^3dx=1/4du)#

#:.x=0=>u=0^4=0 and x=1=>u=1^4=1#

So,

#I=int_0^1 1/(u^2+1)*1/4du#

#=1/4color(red)(int_0^1 1/(u^2+1)du)#

#=1/4[color(red)(tan^-1u)]_0^1#

#=1/4[tan^-1 (1)-tan^-1 (0)]#

#=1/4[pi/4-0]#

#=pi/16#

Note :

#color(red)(int1/(x^2+a^2)dx=1/atan^-1(x/a)+c#

#color(red)(int1/(x^2+1)dx=tan^-1(x)+c,# for , #a=1#