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,

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

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

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

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

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