If \(\int\left(\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}^{3}}\right)\left(\sqrt[23]{3 \mathrm{x}^{-24}+\mathrm{x}^{-26}}\right) \mathrm{dx}\) \[ =-\frac{\alpha}{3(\alpha+1)}\left(3 x^{\beta}+x^{\gamma}\right)^{\frac{\alpha+1}{\alpha}}+C, x>0, \] \((\alpha, \beta, \gamma \in Z)\), where \(C\) is the constant of integration, then \(\alpha+\beta+\gamma\) is equal to ______.

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Indefinite Integrals

\int (\frac{1}{x} + \frac{1}{x ^{3}}) (23 3 x^{- 24} + x^{- 26}) dx

= - \frac{α}{3 ( α + 1 )} (3 x^{β} + x^{γ})^{\frac{α + 1}{α}} + C, x > 0,

(α, β, γ \in Z)

, where

C

is the constant of integration, then

α + β + γ

is equal to ______.

Answer: 19

Since

3 x^{- 24} + x^{- 26} = (3 x^{- 1} + x^{- 3}) x^{- 23}

, for

x > 0

the radical equals

(3 x^{- 1} + x^{- 3})^{1/23} x^{- 1}

, and the integrand becomes

(x^{- 2} + x^{- 4}) (3 x^{- 1} + x^{- 3})^{1/23}

. Put

u = 3 x^{- 1} + x^{- 3}

, so

d u = - 3 (x^{- 2} + x^{- 4}) d x

. The integral is

- \frac{1}{3} \int u^{1/23} d u = - \frac{1}{3} \cdot \frac{u ^{24/23}}{24/23} + C = - \frac{23}{72} (3 x^{- 1} + x^{- 3})^{24/23} + C

. Comparing with

- \frac{α}{3 ( α + 1 )} (3 x^{β} + x^{γ})^{(α + 1) / α} + C

gives

α = 23, β = - 1, γ = - 3

, so

α + β + γ = 19

JEE Main · 2025 · 07 Apr 2025 · Shift 2 · Integer answer