Your integrand is missing some symbols. My best interpretation is the following integral:
[tex]I=\displaystyle\int\frac{x^4+18x^2+4}{x^5+30x^3+20x}\,\mathrm dx[/tex]
Decompose into partial fractions; we're looking for an expansion of the form
[tex]\dfrac{x^4+18x^2+4}{x^5+30x^3+20x}=\dfrac ax+\dfrac{bx^3+cx^2+dx+e}{x^4+30x^2+20}[/tex]
Now:
[tex]x^4+18x^2+4=a(x^4+30x^2+20)+(bx^3+cx^2+dx+e)x[/tex]
[tex]=(a+b)x^4+cx^3+(30a+d)x^2+ex+20a[/tex]
Matching up coefficients tells us that
[tex]\begin{cases}a+b=1\\c=0\\30a+d=18\\e=0\\20a=4\end{cases}\implies a=\dfrac15,b=\dfrac45,d=12[/tex]
so that
[tex]I=\displaystyle\frac15\int\frac{\mathrm dx}x+\frac45\int\frac{x^3+15x}{x^4+30x^2+20}\,\mathrm dx[/tex]
The integral is trivial:
[tex]\displaystyle\frac15\int\frac{\mathrm dx}x=\frac15\ln|x|+C[/tex]
For the second integral, notice that
[tex]\mathrm d(x^4+30x^2+20)=(4x^3+60x)\,\mathrm dx[/tex]
Distribute the 4 over the numerator, then substitute [tex]u=x^4+30x^2+20[/tex] and [tex]\mathrm du=(4x^3+60x)\,\mathrm dx[/tex]:
[tex]\displaystyle\frac15\int\frac{4x^3+60x}{x^4+30x^2+20}\,\mathrm dx=\frac15\int\frac{\mathrm du}u=\frac15\ln|u|+C=\frac15\ln(x^4+30x^2+20)+C[/tex]
So we have
[tex]I=\dfrac15\ln|x|+\dfrac15\ln(x^4+30x^2+20)+C[/tex]
and with some simplification,
[tex]I=\boxed{\ln\sqrt[5]{|x^5+30x^3+20x|}+C}[/tex]
