1. We need to evaluate

$$\lim_{x\to 0}\frac{\alpha x e^x-\beta \ln(1+x)+\gamma x^2 e^{-x}}{x\sin^2 x}=10.$$

Since the denominator tends to $0$, for the limit to be finite, the numerator must vanish up to sufficiently high order.

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2. Expand each term near $x=0$.

We use:

$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots,$$ $$e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\cdots,$$ $$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots,$$ $$\sin x = x-\frac{x^3}{6}+\cdots \implies \sin^2 x = x^2-\frac{x^4}{3}+\cdots$$

Hence,

$$x\sin^2 x = x^3+O(x^5).$$

Now expand the numerator:

$$\alpha x e^x=\alpha x\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots\right) =\alpha x+\alpha x^2+\frac{\alpha}{2}x^3+\frac{\alpha}{6}x^4+\cdots$$ $$-\beta\ln(1+x)=-\beta\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots\right) =-\beta x+\frac{\beta}{2}x^2-\frac{\beta}{3}x^3+\frac{\beta}{4}x^4+\cdots$$ $$\gamma x^2 e^{-x}=\gamma x^2\left(1-x+\frac{x^2}{2}-\frac{x^3}{6}+\cdots\right) =\gamma x^2-\gamma x^3+\frac{\gamma}{2}x^4+\cdots$$

Adding,

$$\text{Numerator}=(\alpha-\beta)x+\left(\alpha+\frac{\beta}{2}+\gamma\right)x^2 +\left(\frac{\alpha}{2}-\frac{\beta}{3}-\gamma\right)x^3+\cdots$$

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3. Since denominator is of order $x^3$, for the limit to be finite we need coefficients of $x$ and $x^2$ in the numerator to be zero.

So,

$$\alpha-\beta=0 \quad\Rightarrow\quad \alpha=\beta$$

and

$$\alpha+\frac{\beta}{2}+\gamma=0.$$

Using $\alpha=\beta$,

$$\alpha+\frac{\alpha}{2}+\gamma=0 \Rightarrow \gamma=-\frac{3\alpha}{2}.$$

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4. Now use the value of the limit.

Because

$$x\sin^2x \sim x^3,$$

the limit equals the coefficient of $x^3$ in the numerator:

$$\frac{\alpha}{2}-\frac{\beta}{3}-\gamma=10.$$

Since $\beta=\alpha$ and $\gamma=-\frac{3\alpha}{2}$,

$$\frac{\alpha}{2}-\frac{\alpha}{3}+\frac{3\alpha}{2}=10.$$

Combine terms:

$$\frac{\alpha}{2}+\frac{3\alpha}{2}=2\alpha,$$

so

$$2\alpha-\frac{\alpha}{3}=10$$ $$\frac{6\alpha-\alpha}{3}=10$$ $$\frac{5\alpha}{3}=10 \Rightarrow \alpha=6.$$

Hence,

$$\beta=6,$$ $$\gamma=-\frac{3\cdot 6}{2}=-9.$$

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5. Therefore,

$$\alpha+\beta+\gamma=6+6-9=3.$$

So the required integer is

$$\boxed{3}.$$

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6. Comparison with stored answer:

Stored correct answer = $3$. Our derived answer also = $3$. Hence, they agree.