If∫(x24+x16+x8)(2x16+3x8+6)1/8dx=1α(2x24+3x16+6x8)βγ+C(where C is constant of integration and β,γare coprime numbers) then the value of (α+β+γ)is

∫ (x23+x15+x7)(2x24+3x16+6x8)1/8dxPut 2x24+3x16+6x8=t148∫t18dt=14889t98+c=154t98+c∴α=54,β=9,γ=8α+β+γ=54+9+8=71

If∫(x24+x16+x8)(2x16+3x8+6)1/8dx=1α(2x24+3x16+6x8)βγ+C(where C is constant of integration and β,γare coprime numbers) then the value of (α+β+γ)is

∫ (x23+x15+x7)(2x24+3x16+6x8)1/8dxPut 2x24+3x16+6x8=t148∫t18dt=14889t98+c=154t98+c∴α=54,β=9,γ=8α+β+γ=54+9+8=71

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If $\int (x^{24} + x^{16} + x^{8}) {(2 x^{16} + 3 x^{8} + 6)}^{1 / 8} d x$ $= \frac{1}{α} {(2 x^{24} + 3 x^{16} + 6 x^{8})}^{\frac{β}{γ}} + C$ (where C is constant of integration and $β, γ$ are coprime numbers) then the value of $(α + β + γ)$ is

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$\int_{}^{} (x^{23} + x^{15} + x^{7}) {(2 x^{24} + 3 x^{16} + 6 x^{8})}^{1 / 8} d x$

Put $2 x^{24} + 3 x^{16} + 6 x^{8} = t$

$\frac{1}{48} \int t^{\frac{1}{8}} d t = \frac{1}{48} (\frac{8}{9}) (t^{\frac{9}{8}}) + c = \frac{1}{54} t^{\frac{9}{8}} + c$

$∴ α = 54, β = 9, γ = 8$

$α + β + γ = 54 + 9 + 8 = 71$

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If∫(x24+x16+x8)(2x16+3x8+6)1/8dx=1α(2x24+3x16+6x8)βγ+C(where C is constant of integration and β,γare coprime numbers) then the value of (α+β+γ)is

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If $\int (x^{24} + x^{16} + x^{8}) {(2 x^{16} + 3 x^{8} + 6)}^{1 / 8} d x$ $= \frac{1}{α} {(2 x^{24} + 3 x^{16} + 6 x^{8})}^{\frac{β}{γ}} + C$ (where C is constant of integration and $β, γ$ are coprime numbers) then the value of $(α + β + γ)$ is