Find the integral ?

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Answer 1 · 2018-06-27T14:19:30

#1/4arcsinh(x^4/2)+C#

#int (x^3*dx)/(sqrt(x^8+4)#

=#1/4int (4x^3*dx)/(sqrt((x^4)^2+4)#

After using #x^4=2sinhu# and #4x^4*dx=2coshu*du# transforms,
this integral became

#1/4int (2coshu*du)/(2coshu)#

=#1/4int du#

=1/4*u+C#

After using #x^4=2sinhu# and #u=arcsinh(x^4/2)# inverse transforms, I found

#1/4arcsinh(x^4/2)+C#

Answer 2 · 2018-06-27T14:28:16

#intx^3/sqrt(4+x^8)dx=1/4ln|x^4+sqrt(x^8+4)|+c#

Here,

#I=intx^3/sqrt(4+x^8)dx=intx^3/sqrt(4+(x^4)^2)dx#

Subst. #x^4=u=>4x^3dx=du=>x^3dx=1/4du#

So,

#I=int1/sqrt(4+u^2)*1/4du#

#=1/4int1/sqrt(u^2+2^2)du#

#=1/4ln|u+sqrt(u^2+2^2)|+c#

#=1/4ln|x^4+sqrt((x^4)^2+2^2)|+c#

#=1/4ln|x^4+sqrt(x^8+4)|+c#