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Indefinite Integrals question

2025 · 7 Apr · Shift 2 · Q48
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Indefinite Integrals question

2025 · 7 Apr · Shift 2 · Q48

JEE MainMathematicsIndefinite IntegralsNumerical+4 / −1
If ∫(1x+1x3)(3x−24+x−2623)dx=−α3(α+1)(3xβ+xγ)α+1α+C,x>0,(α,β,γ∈Z)\int\left(\frac{1}{x}+\frac{1}{x^3}\right)\left(\sqrt[23]{3 x^{-24}+x^{-26}}\right) \mathrm{d} x=-\frac{\alpha}{3(\alpha+1)}\left(3 x^\beta+x^\gamma\right)^{\frac{\alpha+1}{\alpha}}+C, x\gt 0,(\alpha, \beta, \gamma \in \mathbf{Z})∫(x1​+x31​)(233x−24+x−26​)dx=−3(α+1)α​(3xβ+xγ)αα+1​+C,x>0,(α,β,γ∈Z), where C is the constant of integration, then α+β+γ\alpha+\beta+\gammaα+β+γ is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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