Question

Find the integral ?

Answer 1

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

Explanation:

`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

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

Explanation:

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`