The denominator is already factorised. We decompose into x−1C1+x+2C2; with the “cover-up” trick:
C1=5(1+2)2⋅1+3=31,C2=5(−2−1)2(−2)+3=151.
Hence
∫355(x−1)(x+2)2x+3dx=[31ln∣x−1∣+151ln∣x+2∣]35.
Evaluating: (31ln4+151ln7)−(31ln2+151ln5)=31ln2+151ln57≈0,254.