If ∫(1/x + 1/x³) ²³√(3x⁻²⁴ + x⁻²⁶)dx = −α/3(α+1) (3xβ + xγ)α/α+1 + C, where C is the constant of integration, then α + β + γ is equal to:
❓ Question Evaluate the integral: ∫ ( 1 x + 1 x 3 ) 3 x − 24 + x − 26 3 d x = − α 3 ( α + 1 ) ( 3 x β + x γ ) α / ( α + 1 ) + C , where C C is the constant of integration. Find α + β + γ \alpha + \beta + \gamma . 🖼️ Question Image ✍️ Short Solution Factor inside the cube root: 3 x − 24 + x − 26 = x − 26 ( 3 x 2 + 1 ) Then, cube root: 3 x − 24 + x − 26 3 = x − 26 ( 3 x 2 + 1 ) 3 = x − 26 / 3 ( 3 x 2 + 1 ) 1 / 3 . Simplify the squared bracket: ( 1 x + 1 x 3 ) = x 2 + 1 x 3 . \left(\frac{1}{x} + \frac{1}{x^3}\right) = \frac{x^2+1}{x^3}. Raise to first power (as given), then multiply with cube root: ( 1 x + 1 x 3 ) 3 x − 24 + x − 26 3 = x 2 + 1 x 3 ⋅ x − 26 / 3 ( 3 x 2 + 1 ) 1 / 3 = ( x 2 + 1 ) x − 3 − 26 / 3 ( 3 x 2 + 1 ) 1 / 3 . \left(\frac{1}{x} + \frac{1}{x^3}\right) \sqrt[3]{3x^{-24} + x^{-26}} = \frac{x^2+1}{x^3} \cdot x^{-26/3} (3x^2 +1)^{1/3} = (x^2+1) x^{-3-26/3} (3x^2+1)^{1/3}. Compute exponent: − 3 − 26 / 3 = − 35 / 3 -3 - 26/3 = -35/3 . So: (...