Partial fractions: (x+1)(x+2)x=x+1−1+x+22.
∫12[x+1−1+x+22]dx=[−ln∣x+1∣+2ln∣x+2∣]12
=(−ln3+2ln4)−(−ln2+2ln3)=ln2732.
∫12(x+1)(x+2)xdx=
Held on 22 May 2023 · Verified 13 Jul 2026.
tan−12732
tan−13227
log2732
log3227
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