Exams Joint Entrance Examination Mathematics Indefinite Integrals Question 24 Question 24 of 416 Medium NUMERICAL
If ∫ ( 1 x + 1 x 3 ) ( 233 x − 24 + x − 26 ) d x = − α 3 ( α + 1 ) ( 3 x β + x γ ) α + 1 α + C , x > 0 , ( α , β , γ ∈ Z ) \int (\frac{1}{x}+\frac{1}{{x}^{3}})(\sqrt{233}{x}^{-24}+{x}^{-26})dx=-\frac{\alpha }{3(\alpha +1)}{(3{x}^{\beta }+{x}^{\gamma })}^{\frac{\alpha +1}{\alpha }}+C,x>0,(\alpha ,\beta ,\gamma \in Z) ∫ ( x 1 + x 3 1 ) ( 233 x − 24 + x − 26 ) d x = − 3 ( α + 1 ) α ( 3 x β + x γ ) α α + 1 + C , x > 0 , ( α , β , γ ∈ Z ) , where C is the constant of integration, then α + β + γ \alpha +\beta +\gamma α + β + γ is equal to \_\_\_\_\_\_\_\_\_\_\_.
Answer: 19
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∫ ( 1 x 2 + 1 x 4 ) ( 3 x + 1 x 3 ) 1 23 d x t = 3 x + 1 x 3 ⇒ d t = − 3 ( 1 x 2 + 1 x 4 ) d x ∫ t 1 / 23 d t − 3 = t 24 / 23 ( 24 23 ) ( − 3 ) + C ⇒ α = 23 β = − 1 γ = − 3 α + β + γ = 19 \begin{gathered} \int (\frac{1}{{x}^{2}}+\frac{1}{{x}^{4}}){(\frac{3}{x}+\frac{1}{{x}^{3}})}^{\frac{1}{23}}dx \\ t=\frac{3}{x}+\frac{1}{{x}^{3}}\Rightarrow dt=-3(\frac{1}{{x}^{2}}+\frac{1}{{x}^{4}})dx \\ \int \frac{{t}^{1/23}dt}{-3}=\frac{{t}^{24/23}}{(\frac{24}{23})(-3)}+C \\ \Rightarrow \alpha =23\beta =-1\gamma =-3 \\ \alpha +\beta +\gamma =19 \end{gathered} ∫ ( x 2 1 + x 4 1 ) ( x 3 + x 3 1 ) 23 1 d x t = x 3 + x 3 1 ⇒ d t = − 3 ( x 2 1 + x 4 1 ) d x ∫ − 3 t 1/23 d t = ( 23 24 ) ( − 3 ) t 24/23 + C ⇒ α = 23 β = − 1 γ = − 3 α + β + γ = 19 Answer the question to unlock