We need
#intx^ndx=x^(n+1)/(n+1)+C(n!=-1)#
We perform a polynomial long division
#color(white)(aaaa)##x^4##color(white)(aaaaaaaaa)##∣##x^2+1#
#color(white)(aaaa)##x^4+x^2##color(white)(aaaaa)##∣##x^2-1#
#color(white)(aaaaa)##0-x^2#
#color(white)(aaaaaaa)##-x^2-1#
#color(white)(aaaaaaaaa)##0+1#
Therefore,
#x^4/(x^2+1)=(x^2-1)+1/(x^2+1)#
So,
#int(x^4dx)/(x^2+1)=int(x^2-1)dx+color(red)(intdx/(x^2+1))#
#=x^3/3-x+color(red)(intdx/(x^2+1))#
The integral in red is done by trigonometric substitution
Let #x=tantheta#, #=>#, #dx=sec^2theta d theta#
and, #x^2+1=tan^2theta+1=sec^2theta#
Therefore,
#color(red)(intdx/(x^2+1)=int(sec^2theta d theta)/sec^2theta=intd theta=theta=arctan(x))#
Putting it all together
#int(x^4dx)/(x^2+1)=x^3/3-x+arctan(x)+C#
