If ∫ (1/x + 1/x³) (3x⁻²⁴ + x⁻²⁶)^(1/23) dx = -α/(3(α+1)) (3xᵝ + xᵞ)^((α+1)/α) + C, x 0, (α, β, γ ∈

Q: Wait, why do we factor out $x^{-23}$ specifically? Could we have factored out something else?

Great question! We factor out $x^{-23}$ because the exponent outside the bracket is $\frac{1}{23}$. When we apply the power rule $(ab)^n = a^n b^n$, the term $x^{-23}$ becomes $(x^{-23})^{\frac{1}{23}} = x^{-1}$. This $x^{-1}$ is crucial because it helps us transform the rest of the integral into a form that matches the derivative of our substitution. If we chose any other power, the resulting expression would not simplify into a form that allows for easy substitution.

Q: I am confused about the derivative. How did $3x^{-1} + x^{-3}$ become $-3(x^{-2} + x^{-4})$?

Let's break it down using the power rule for differentiation: $\frac{d}{dx}(x^n) = nx^{n-1}$. For $3x^{-1}$, the derivative is $3(-1)x^{-2} = -3x^{-2}$. For $x^{-3}$, the derivative is $(-3)x^{-4}$. So, the derivative of $3x^{-1} + x^{-3}$ is $-3x^{-2} - 3x^{-4}$. If you factor out the $-3$, you get $-3(x^{-2} + x^{-4})$. It is a common place to make a sign error, so always double-check your negative exponents!

Q: The target form has $\alpha$ in the denominator and the exponent. How do I match my result to that form?

This is a test of pattern recognition. After integrating, you will have a result like $-\frac{23}{3(24)} u^{\frac{24}{23}}$. The target form is $-\frac{\alpha}{3(\alpha+1)} (3x^\beta + x^\gamma)^{\frac{\alpha+1}{\alpha}}$. By comparing the components, you can see that $\alpha = 23$ makes the fraction $-\frac{23}{3(23+1)} = -\frac{23}{3(24)}$ and the exponent $\frac{23+1}{23} = \frac{24}{23}$. It is all about aligning the constants and powers to reveal the hidden values.

Question

If $\int (\frac{1}{x} + \frac{1}{x^3}) (3x^{-24} + x^{-26})^{1/23} dx = -\frac{\alpha}{3(\alpha+1)} (3x^\beta + x^\gamma)^{\frac{\alpha+1}{\alpha}} + C, x > 0$, ($\alpha, \beta, \gamma \in \mathbb{Z}$), where $C$ is the constant of integration, then $\alpha + \beta + \gamma$ is equal to ________ .

Sigma Percentile · Accepted Answer

### The Mountain of Calculus: A Journey Through the Integral

Imagine standing before a massive, jagged mountain. That is exactly what this integral looks like at first glance:

$$I = \int \left( \frac{1}{x} + \frac{1}{x^3} \right) (3x^{-24} + x^{-26})^{\frac{1}{23}} dx$$

It is intimidating, filled with negative powers and a fractional exponent that seems designed to confuse. But in the world of JEE Advanced, every complex problem is just a series of simple, elegant steps waiting to be uncovered. Let us climb this mountain together.

### Phase 1: The Art of Factoring

The secret to conquering this integral lies in the term $(3x^{-24} + x^{-26})^{\frac{1}{23}}$. We need to simplify this. The intuition here is to factor out a power of $x$ from inside the bracket that will perfectly cancel with the $\frac{1}{23}$ exponent.

If we factor out $x^{-23}$, we get:

$$(3x^{-24} + x^{-26})^{\frac{1}{23}} = [x^{-23} (3x^{-1} + x^{-3})]^{\frac{1}{23}}$$

Now, applying the property $(ab)^n = a^n b^n$, we get $(x^{-23})^{\frac{1}{23}} (3x^{-1} + x^{-3})^{\frac{1}{23}}$, which simplifies beautifully to $x^{-1} (3x^{-1} + x^{-3})^{\frac{1}{23}}$. This is the breakthrough!

### Phase 2: The Bridge to Simplicity

Now, let us bring this back to our integral. The expression becomes:

$$I = \int \left( \frac{1}{x} + \frac{1}{x^3} \right) x^{-1} (3x^{-1} + x^{-3})^{\frac{1}{23}} dx$$

If we distribute the $x^{-1}$ into the first bracket $(\frac{1}{x} + \frac{1}{x^3})$, we get $(x^{-2} + x^{-4})$. Our integral is now:

$$I = \int (x^{-2} + x^{-4}) (3x^{-1} + x^{-3})^{\frac{1}{23}} dx$$

Do you see the beauty of this? The term outside the bracket is almost exactly the derivative of the term inside the bracket. This is the classic setup for the method of substitution.

### Phase 3: The Substitution

Let us set $u = 3x^{-1} + x^{-3}$. Now, we differentiate with respect to $x$:

$$\frac{du}{dx} = 3(-1)x^{-2} + (-3)x^{-4} = -3(x^{-2} + x^{-4})$$

This means $(x^{-2} + x^{-4}) dx = -\frac{1}{3} du$. Substituting this into our integral, we get:

$$I = \int u^{\frac{1}{23}} \left( -\frac{1}{3} \right) du = -\frac{1}{3} \int u^{\frac{1}{23}} du$$

The mountain has suddenly become a gentle hill. We can now use the standard power rule for integration: $\int u^n du = \frac{u^{n+1}}{n+1} + C$. Adding one to the power $\frac{1}{23}$ gives us $\frac{24}{23}$.

So, the integral is:

$$I = -\frac{1}{3} \left[ \frac{u^{\frac{24}{23}}}{\frac{24}{23}} \right] + C = -\frac{1}{3} \cdot \frac{23}{24} u^{\frac{24}{23}} + C = -\frac{23}{3(24)} u^{\frac{24}{23}} + C$$

### Phase 4: The Final Victory

We are almost at the summit. Substituting $u = 3x^{-1} + x^{-3}$ back into our result, we get:

$$I = -\frac{23}{3(24)} (3x^{-1} + x^{-3})^{\frac{24}{23}} + C$$

Now, we compare this to the target form: $-\frac{\alpha}{3(\alpha+1)} (3x^\beta + x^\gamma)^{\frac{\alpha+1}{\alpha}} + C$. By direct comparison, we identify $\alpha = 23$, $\beta = -1$, and $\gamma = -3$.

The final step is just simple arithmetic:

$$\alpha + \beta + \gamma = 23 + (-1) + (-3) = 19$$

We have reached the top! The complex integral has been solved, and the answer is **19**. Remember, no matter how terrifying a problem looks, there is always a path to the solution if you break it down step by step.

Enter Numerical Value:

Visualized Solution

Analyze the Integral Structure

The Factoring Strategy

Simplify the Power Term

Merge the Terms

The New Integral Form

Identify the Substitution

Calculate the Differential

Substitute into the Integral

Perform the Integration

Simplify the Result

Compare with the Given Form

Final Calculation

The Sigma Insight: Integration by Substitution

The Mountain of Calculus

Phase 1

Phase 2

Phase 3

Phase 4

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