If aligned ∫((1)/(x)+(1)/(x^3)) & ( [23] 3 x^ -24 +x^ -26 ) d x & =-(α)/(3(α+1))(3 x^β+x^γ)^ (α+1)/(α) +C, x 0, aligned (α, β, γ ∈ Z) , where C is the constant of integration, then α+β+γ is equal to ________ .

Correct answer: 0 aligned & ∫( 1 x^2 + 1 x^4 )( 3 x + 1 x^3 )^ (1)/(23) dx & using t= 3 x + 1 x^3 dt=-3( 1 x^2 + 1 x^4 ) dx & ∫ t^ 1 / 23 dt -3 = t^ 24 / 23 ((24)/(23))(-3) +C & α=23 β=-1 γ=-3 & α+β+γ=19 aligned

JEE Main 20257 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 7 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2025 · 7 April · Evening Shift · Mathematics · Indefinite Integration

If aligned ∫((1)/(x)+(1)/(x^3)) & ( [23] 3 x^ -24 +x^ -26 ) d x &...

\begin{aligned} \int\left(\frac{1}{x}+\frac{1}{x^3}\right) & \left(\sqrt[23]{3 x^{-24}+x^{-26}}\right) d x \\ & =-\frac{\alpha}{3(\alpha+1)}\left(3 x^\beta+x^\gamma\right)^{\frac{\alpha+1}{\alpha}}+C, x \gt 0,\end{aligned}

(\alpha, \beta, \gamma \in Z)

, where

is the constant of integration, then

\alpha+\beta+\gamma$ is equal to ________ .

Correct answer

Step-by-step solution

\begin{aligned} & \int\left(\frac{1}{\mathrm{x}^2}+\frac{1}{\mathrm{x}^4}\right)\left(\frac{3}{\mathrm{x}}+\frac{1}{\mathrm{x}^3}\right)^{\frac{1}{23}} \mathrm{dx} \\ & \text{ using } \mathrm{t}=\frac{3}{\mathrm{x}}+\frac{1}{\mathrm{x}^3} \Rightarrow \mathrm{dt}=-3\left(\frac{1}{\mathrm{x}^2}+\frac{1}{\mathrm{x}^4}\right) \mathrm{dx} \\ & \int \frac{\mathrm{t}^{1 / 23} \mathrm{dt}}{-3}=\frac{\mathrm{t}^{24 / 23}}{\left(\frac{24}{23}\right)(-3)}+C \\ & \Rightarrow \alpha=23 \beta=-1 \gamma=-3 \\ & \alpha+\beta+\gamma=19\end{aligned}

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