If ∫(1x+1x3)(3x–24+x–2623)dx=–α3(α+1)(3xβ+xγ)α+1α+C, x > 0, (α,β,γ∈Z), where c is the constant of integration, then α+β+γ is equal to __________. [2025]
(19)
We have, ∫(1x2+1x4)(3x+1x3)123dx
Put t=3x+1x3 ⇒ dt=–3(1x2+1x4)dx
Now, ∫t1/23dt–3=t24/23(2423)(–3)+C
On comparing, we get α=23, β=–1, γ=–3
∴ α+β+γ=19.