[tex]x^3-125=(x-5)(x^2+5x+25)[/tex]
[tex]\dfrac{17}{(x-5)(x^2+5x+25)}=\dfrac a{x-5}+\dfrac{bx+c}{x^2+5x+25}[/tex]
[tex]\dfrac{17}{(x-5)(x^2+5x+25)}=\dfrac{a(x^2+5x+25)+(bx+c)(x-5)}{(x-5)(x^2+5x+25)}[/tex]
[tex]\implies17=(a+b)x^2+(5a-5b+c)x+(25a-5c)[/tex]
[tex]\implies\begin{cases}a+b=0\\5a-5b+c=0\\25a-5c=17\end{cases}\implies a=\dfrac{17}{75},b=-\dfrac{17}{75},c=-\dfrac{34}{15}[/tex]
[tex]\displaystyle\int\dfrac{17}{x^3-125}\,\mathrm dx=\dfrac{17}{75}\int\dfrac{\mathrm dx}{x-5}-\dfrac{17}{75}\int\frac x{x^2+5x+25}\,\mathrm dx-\dfrac{34}{15}\int\dfrac{\mathrm dx}{x^2+5x+25}[/tex]
[tex]\displaystyle=\dfrac{17}{75}\int\dfrac{\mathrm dx}{x-5}-\dfrac{17}{150}\int\frac{2x+5}{x^2+5x+25}\,\mathrm dx-\dfrac{17}{10}\int\dfrac{\mathrm dx}{x^2+5x+25}[/tex]
[tex]\displaystyle=\dfrac{17}{75}\int\dfrac{\mathrm dx}{x-5}-\dfrac{17}{150}\int\frac{2x+5}{x^2+5x+25}\,\mathrm dx-\dfrac{17}{10}\int\dfrac{\mathrm dx}{\left(x+\frac52\right)^2+\frac{75}4}[/tex]
The first integral is trivial. For the second, replace [tex]y=x^2+5x+25[/tex] so that [tex]\mathrm dy=(2x+5)\,\mathrm dx[/tex]. For the third, take [tex]x+\dfrac52=\dfrac{5\sqrt3}2\tan z[/tex] so that [tex]\mathrm dx=\dfrac{5\sqrt3}2\sec^2z\,\mathrm dz[/tex].
[tex]\displaystyle=\dfrac{17}{75}\ln|x-5|-\dfrac{17}{150}\int\frac{\mathrm dy}y-\dfrac{17}{10}\int\dfrac{\frac{5\sqrt3}2\sec^2z}{\left(\frac{5\sqrt3}2\tan z\right)^2+\frac{475}4}\,\mathrm dz[/tex]
[tex]\displaystyle=\dfrac{17}{75}\ln|x-5|-\dfrac{17}{150}\ln|y|-\dfrac{17}{25\sqrt3}\int\dfrac{\sec^2z}{\tan^2z+1}\,\mathrm dz[/tex]
[tex]\displaystyle=\dfrac{17}{75}\ln|x-5|-\dfrac{17}{150}\ln(x^2+5x+25)-\dfrac{17}{25\sqrt3}\int\mathrm dz[/tex]
[tex]\displaystyle=\dfrac{17}{75}\ln|x-5|-\dfrac{17}{150}\ln(x^2+5x+25)-\dfrac{17}{25\sqrt3}z+C[/tex]
[tex]\displaystyle=\dfrac{17}{75}\ln|x-5|-\dfrac{17}{150}\ln(x^2+5x+25)-\dfrac{17}{25\sqrt3}\arctan\dfrac{2x+5}{5\sqrt3}+C[/tex]
