Expanding: (x+x2−1)2=x2+(x2−1)+2xx2−1=2x2−1+2xx2−1. Integrating term by term: ∫2x2dx=32x3, ∫−1dx=−x, ∫2xx2−1dx=32(x2−1)3/2. So α=−1, β=2/3, γ=2/3. Then 3(α+β+γ)=3(−1+4/3)=1.
If ∫(x+x2−1)2dx=α⋅x+βx3+γ(x2−1)23+C, where C is arbitrary constant, then the value of 3(α+β+γ) is
Held on 10 Aug 2022 · Verified 13 Jul 2026.
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