We need
#intx^ndx=x^(n+1)/(n+1)+C(n!=-1)#
We factorise the denominator
#x^4+x^2=x^2(x^2+1)#
Now we can do the decomposition into partial fractions
#1/(x^4+x^2)=A/x^2+B/x+(Cx+D)/(x^2+1)#
#=(A(x^2+1)+Bx(x^2+1)+(Cx+D)(x^2))/(x^2(x^2+1))#
Therefore,
#1=A(x^2+1)+Bx(x^2+1)+(Cx+D)(x^2)#
Let, #x=0#,#=>#,#1=A#
Coefficients of #x^2#, #=>#, #0=A+D#, #=>#, #D=-1#
Coefficients of #x#, #=>#, #0=B#
Coeficients of #x^3#, #=>#, #0=B+C#, #=>#, #C=0#
So,
#1/(x^4+x^2)=1/x^2+0/x+(0x-1)/(x^2+1)#
#=1/x^2-1/(x^2+1)#
Therefore,
#intdx/(x^4+x^2)=intdx/x^2-intdx/(x^2+1)#
The first integral is #intdx/x^2=-1/x#
The second integral is #intdx/(x^2+1)#
We use a trigonometric substitution
Let #x=tanu#, #=>#, #du=sec^2udu#
and #x^2+1=tan^2u+1=sec^2u#
So,
#intdx/(x^2+1)=int(sec^2udu)/sec^2u=intdu=u#
#=arctanx#
Putting it alltogether,
#intdx/(x^4+x^2)=-1/x-arctanx+C#
