If ∫ (1/x + 1/x^3) (23√(3x^(-24) + x^(-26))) dx = α/3(α+1) (3x^β + x^γ)^(α+1/α) + C, x>0, (α, β, γ ∈ Z),

Question

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 _____.

Answer 1

Answer is: 19 

\(\int\left(\frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{x}^{4}}\right)\left(\frac{3}{\mathrm{x}}+\frac{1}{\mathrm{x}^{3}}\right)^{\frac{1}{23}} \mathrm{dx}\)

using \(\mathrm{t}=\frac{3}{\mathrm{x}}+\frac{1}{\mathrm{x}^{3}} \Rightarrow \mathrm{dt}=-3\left(\frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{x}^{4}}\right) \mathrm{dx}\)

\(\int \frac{\mathrm{t}^{1 / 23} \mathrm{dt}}{-3}=\frac{\mathrm{t}^{24 / 23}}{\left(\frac{24}{23}\right)(-3)}+\mathrm{C}\)

\(\Rightarrow \alpha=23 \beta=-1 \gamma=-3\)

\(\alpha+\beta+\gamma=19\)