Find Alpha Beta Gamma from Integral Expression

 ❓ Question

Evaluate the integral:

$$\int (\frac{1}{x}+\frac{1}{x^3})\sqrt[3]{3x^{-24}+x^{-26}}dx=-\frac{\alpha }{3(\alpha +1)}(3x^{\beta }+x^{\gamma }{)}^{\alpha \mathrm{/}(\alpha +1)}+C,$$

where $C$ is the constant of integration. Find $\alpha + \beta + \gamma$.

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Substitution method
👉 Reverse chain rule
👉 Standard integration pattern.

Main idea:

Look for form:

$$f'(x)\,[f(x)]^n$$

whose integral is:

$$\frac{[f(x)]^{n+1}}{n+1}$$

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🔷 Step 1 — Rewrite Integrand 💯

Given:

$$\int \left(\frac1x+\frac1{x^3}\right) \left(3x^{-24}+x^{-26}\right)^{\frac1{23}} dx$$

Observe:

$$3x^{-24}+x^{-26} = x^{-26}(3x^2+1)$$

But better approach is direct differentiation.

Let:

$$u=3x^{-24}+x^{-26}$$

Then:

$$\frac{du}{dx} = -72x^{-25}-26x^{-27}$$

Factor:

$$= -2x^{-27}(36x^2+13)$$

Now rewrite carefully from original expression.

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🔷 Step 2 — Spot Better Substitution

Notice:

$$3x^{-24}+x^{-26} = x^{-26}(3x^2+1)$$

and:

$$\frac1x+\frac1{x^3} = \frac{x^2+1}{x^3}$$

The intended form matches:

$$\int f'(x)\,[f(x)]^{1/23}dx$$

Let:

$$u=3x^\beta+x^\gamma$$

From expression:

$$u=3x^{-24}+x^{-26}$$

Thus:

$$\beta=-24,\qquad \gamma=-26$$

Exponent outside is:

$$\frac{\alpha+1}{\alpha}$$

Since integrand power is:

$$\frac1{23}$$

we compare with standard result:

$$\int u^{1/23}du = \frac{23}{24}u^{24/23}$$

Hence:

$$\frac{\alpha+1}{\alpha} = \frac{24}{23}$$

Thus:

$$\alpha=23$$

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🔷 Step 3 — Find Required Sum

$$\alpha+\beta+\gamma = 23+(-24)+(-26)$$
$$=23-50$$
$$=-27$$

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🔷 Step 4 — JEE Trap Alert 🚨

❌ Power comparison galat kar dena

❌ Outer exponent directly $1/23$ maan lena

❌ $\beta,\gamma$ signs miss kar dena

Remember:

$$\boxed{ \int u^n\,du=\frac{u^{n+1}}{n+1} }$$

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✅ Final Answer

$$\boxed{{\displaystyle -27}}$$

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