1. Given integral

We need to evaluate

$$I=\int \left(\frac1x+\frac1{x^3}\right)\left(\sqrt[23]{3x^{-24}+x^{-26}}\right)\,dx.$$

Since

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

we rewrite:

$$I=\int \left(\frac1x+\frac1{x^3}\right)(3x^{-24}+x^{-26})^{1/23}\,dx.$$

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2. Look for substitution

Notice

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

has derivative

$$\frac{d}{dx}(3x^{-24}+x^{-26})=-72x^{-25}-26x^{-27}.$$

Factor this:

$$-72x^{-25}-26x^{-27}=-2\left(36x^{-25}+13x^{-27}\right).$$

That does not directly match the factor $\left(\frac1x+\frac1{x^3}\right)$, so let us instead inspect the answer form.

We are given that

$$I=-\frac{\alpha}{3(\alpha+1)}\left(3x^\beta+x^\gamma\right)^{\frac{\alpha+1}{\alpha}}+C.$$

This suggests the integrand is of the form

$$(3x^\beta+x^\gamma)^{1/\alpha}\cdot \frac{d}{dx}(3x^\beta+x^\gamma).$$

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3. Match the inner expression

The given integrand contains

$$(3x^{-24}+x^{-26})^{1/23}.$$

So naturally we identify

$$\alpha=23,$$

and

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

Now verify whether the outside factor matches the derivative pattern.

Let

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

Then

$$\frac{d\nu}{dx}=-72x^{-25}-26x^{-27}=-2x^{-27}(36x^2+13).$$

But the factor in the integrand is

$$\frac1x+\frac1{x^3}=x^{-1}+x^{-3}=x^{-3}(x^2+1).$$

So direct substitution in $u=3x^{-24}+x^{-26}$ does not fit immediately.

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4. Try factoring the inner expression

Observe:

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

Hence

$$(3x^{-24}+x^{-26})^{1/23}=x^{-26/23}(3x^2+1)^{1/23}.$$

Therefore the integrand becomes

$$\left(x^{-1}+x^{-3}\right)x^{-26/23}(3x^2+1)^{1/23}.$$

This is not especially helpful directly.

So instead let us differentiate the given answer form and compare.

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5. Differentiate the proposed antiderivative form

Suppose

$$I=-\frac{\alpha}{3(\alpha+1)}(3x^\beta+x^\gamma)^{\frac{\alpha+1}{\alpha}}+C.$$

Differentiating,

$$I'= -\frac{\alpha}{3(\alpha+1)}\cdot \frac{\alpha+1}{\alpha}(3x^\beta+x^\gamma)^{1/\alpha}(3\beta x^{\beta-1}+\gamma x^{\gamma-1}).$$

So

$$I'=-\frac13 (3x^\beta+x^\gamma)^{1/\alpha}(3\beta x^{\beta-1}+\gamma x^{\gamma-1}).$$

This must equal

$$\left(\frac1x+\frac1{x^3}\right)(3x^{-24}+x^{-26})^{1/23}.$$

Thus we must have

$$\alpha=23,$$

and

$$3x^\beta+x^\gamma=3x^{-24}+x^{-26}.$$

Hence

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

Then

$$3\beta x^{\beta-1}+\gamma x^{\gamma-1}=3(-24)x^{-25}+(-26)x^{-27}=-72x^{-25}-26x^{-27}.$$

Therefore

$$I'=-\frac13(3x^{-24}+x^{-26})^{1/23}(-72x^{-25}-26x^{-27}).$$ $$I'=(3x^{-24}+x^{-26})^{1/23}\left(24x^{-25}+\frac{26}{3}x^{-27}\right).$$

This does not match the integrand exactly as typed.

So there is likely a typo in the printed coefficient form, but the intended identification from the structure is clearly

$$\alpha=23,\quad \beta=-24,\quad \gamma=-26.$$

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6. Compute the required sum

$$\alpha+\beta+\gamma=23-24-26=-27.$$

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7. Compare with stored answer

Stored correct answer is $19$, but the expression given in the question leads to

$$\boxed{-27}.$$

So I do not agree with the stored answer.

The most plausible issue is a typo in the question statement or answer key.