If , x 0, , where c is the constant of integration, then is equal to __________. [2025]

Question

If $\int (\frac{1}{x} + \frac{1}{x^{3}}) (\sqrt[23]{3 x^{- 24} + x^{- 26}}) d x = - \frac{α}{3 (α + 1)} {(3 x^{β} + x^{γ})}^{\frac{α + 1}{α}} + C$ , x > 0, $(α, β, γ \in Z)$ , where c is the constant of integration, then $α + β + γ$ is equal to __________. [2025]

Smart Achievers · Accepted Answer

(19)

We have, $\int (\frac{1}{x^{2}} + \frac{1}{x^{4}}) {(\frac{3}{x} + \frac{1}{x^{3}})}^{\frac{1}{23}} d x$

Put $t = \frac{3}{x} + \frac{1}{x^{3}} \Rightarrow d t = - 3 (\frac{1}{x^{2}} + \frac{1}{x^{4}}) d x$

Now, $\int \frac{t^{1 / 23} d t}{- 3} = \frac{t^{24 / 23}}{(\frac{24}{23}) (- 3)} + C$

On comparing, we get $α = 23, β = - 1, γ = - 3$

$∴ α + β + γ = 19$ .

Solution

Q.
If $\int (\frac{1}{x} + \frac{1}{x^{3}}) (\sqrt[23]{3 x^{- 24} + x^{- 26}}) d x = - \frac{α}{3 (α + 1)} {(3 x^{β} + x^{γ})}^{\frac{α + 1}{α}} + C$ , x > 0, $(α, β, γ \in Z)$ , where c is the constant of integration, then $α + β + γ$ is equal to __________. [2025]

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Solution

Q. 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]

Ans. (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.

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Q.
If $\int (\frac{1}{x} + \frac{1}{x^{3}}) (\sqrt[23]{3 x^{- 24} + x^{- 26}}) d x = - \frac{α}{3 (α + 1)} {(3 x^{β} + x^{γ})}^{\frac{α + 1}{α}} + C$ , x > 0, $(α, β, γ \in Z)$ , where c is the constant of integration, then $α + β + γ$ is equal to __________. [2025]