Let I=∫dxx(x3+8)=13∫3x2dxx3(x3+8)
Put x3+8=t⇒3x3dx=dt
I=13∫dt(t−8)t⇒13∫18×(1t−8−1t)
⇒124[log|t−8|−log|t|]+C
I=124log∣∣∣x3x3+8∣∣∣+C