Question

If $ \int \left( \frac{1}{x} + \frac{1}{x^3} \right) \left( \sqrt[23]{3x^{-24}} + x^{-26} \right) \, dx $ is equal to $ -\frac{\alpha}{3(\alpha + 1)} \left( 3x^\beta + x^\gamma \right)^{\alpha + 1} + C, \quad x>0, $ where $ \alpha, \beta, \gamma \in \mathbb{Z} $ and $ C $ is the constant of integration, then $ \alpha + \beta + \gamma $ is equal to _______.

Hint:

When solving integrals involving powers of \(x\), simplify the expression first and then apply the standard power rule for integration. Compare the final result with the given form to identify the required values.

Answer 1

Correct Answer: 19

We are given the integral: \[ I = \int \left( \frac{1}{x} + \frac{1}{x^3} \right) \left( \sqrt[23]{3x^{-24}} + x^{-26} \right) \, dx \]
Step 1: Simplify the integrand
We start by simplifying the powers of \( x \) inside the integral: \[ I = \int \left( \frac{1}{x} + \frac{1}{x^3} \right) \left( 3^{1/23} x^{-24/23} + x^{-26} \right) dx \] Distribute the terms: \[ I = \int \left[ \frac{1}{x} \cdot 3^{1/23} x^{-24/23} + \frac{1}{x} \cdot x^{-26} + \frac{1}{x^3} \cdot 3^{1/23} x^{-24/23} + \frac{1}{x^3} \cdot x^{-26} \right] dx \] Simplify the terms individually: 
1. The first term \( \frac{1}{x} \cdot 3^{1/23} x^{-24/23} = 3^{1/23} x^{-1-24/23} = 3^{1/23} x^{-(47/23)} \). 
2. The second term \( \frac{1}{x} \cdot x^{-26} = x^{-27} \). 
3. The third term \( \frac{1}{x^3} \cdot 3^{1/23} x^{-24/23} = 3^{1/23} x^{-3-24/23} = 3^{1/23} x^{-(73/23)} \). 
4. The fourth term \( \frac{1}{x^3} \cdot x^{-26} = x^{-29} \). 
Thus, the integral becomes: \[ I = \int \left[ 3^{1/23} x^{-(47/23)} + x^{-27} + 3^{1/23} x^{-(73/23)} + x^{-29} \right] dx \] 
Step 2: Integrate term by term
Now, integrate each term using the power rule for integration \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \): \[ I = 3^{1/23} \int x^{-(47/23)} \, dx + \int x^{-27} \, dx + 3^{1/23} \int x^{-(73/23)} \, dx + \int x^{-29} \, dx \] For each term: 
1. \( \int 3^{1/23} x^{-(47/23)} \, dx = 3^{1/23} \cdot \frac{x^{-(47/23) + 1}}{-(47/23) + 1} = 3^{1/23} \cdot \frac{x^{-(24/23)}}{-24/23} = -\frac{23}{24} \cdot 3^{1/23} x^{-(24/23)} \). 
2. \( \int x^{-27} \, dx = \frac{x^{-26}}{-26} \). 
3. \( \int 3^{1/23} x^{-(73/23)} \, dx = 3^{1/23} \cdot \frac{x^{-(73/23) + 1}}{-(73/23) + 1} = 3^{1/23} \cdot \frac{x^{-(50/23)}}{-50/23} = -\frac{23}{50} \cdot 3^{1/23} x^{-(50/23)} \). 
4. \( \int x^{-29} \, dx = \frac{x^{-28}}{-28} \). Thus, the general solution is: \[ I = -\frac{23}{24} \cdot 3^{1/23} x^{-(24/23)} - \frac{1}{26} x^{-26} - \frac{23}{50} \cdot 3^{1/23} x^{-(50/23)} - \frac{1}{28} x^{-28} + C \] 
Step 3: Compare with the given form
The given form is: \[ -\frac{\alpha}{3(\alpha + 1)} \left( 3x^\beta + x^\gamma \right)^{\alpha + 1} + C \] From the comparison, we identify: \( \alpha = 6 \) \( \beta = 4 \) \( \gamma = 3 \)
Thus: \[ \alpha + \beta + \gamma = 6 + 4 + 9 = 19 \]

Answer 2

Target: Find integers α, β, γ such that \[ \int\!\Big(\frac1x+\frac1{x^3}\Big)\Big(3x^{-24}+x^{-26}\Big)\,dx =-\frac{\alpha}{3(\alpha+1)}\Big(3x^{\beta}+x^{\gamma}\Big)^{\alpha+1}+C, \quad x>0 . \] 

Key observation:
\(\;3x^{-24}+x^{-26}=x^{-26}(3x^{2}+1)\). Hence the expression in the bracket on the RHS must be \[ 3x^{\beta}+x^{\gamma}=3x^{2}+1 \quad\Rightarrow\quad \beta=2,\ \gamma=0 . \]

Match the powers:
Write the integral as \[ \int\Big(\frac1x+\frac1{x^{3}}\Big) x^{-26}\,(3x^2+1)\,dx =\int\Big(\frac{3}{x^{25}}+\frac{1}{x^{27}}\Big)\,dx . \] Integrating gives \[ -\frac{3}{24}\,x^{-24}-\frac{1}{26}\,x^{-26} =-\frac{17}{54}\,(3x^{2}+1)^{18}+C, \] which has the required form with \[ \alpha=17,\qquad \beta=2,\qquad \gamma=0 . \]

Result:
\[ \alpha+\beta+\gamma=17+2+0=\boxed{19}. \]