1. Given function

$$f(t)=\int_0^t e^{x^3}\left(\frac{x^8}{(x^6+2x^3+2)^2}\right)\,dx$$

We need to find $f(1)+f'(1)$ and compare it with

$$\alpha e-\frac16.$$

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2. Simplify the integrand

Notice that

$$x^6+2x^3+2=(x^3+1)^2+1.$$

Let

$$u=x^3 \implies d\nu=3x^2dx.$$

But the given integrand has $x^8dx=x^6x^2dx=(x^3)^2x^2dx=\nu^2x^2dx$. So,

$$x^8dx=\nu^2\frac{d\nu}{3}.$$

Also,

$$e^{x^3}=e^{\nu}, \qquad x^6+2x^3+2=\nu^2+2\nu+2.$$

Hence

$$f(t)=\frac13\int_0^{t^3} \frac{e^{\nu}\nu^2}{(\nu^2+2\nu+2)^2}\,d\nu.$$

Now observe:

$$\frac{d}{d\nu}\left(\frac{e^{\nu}}{\nu^2+2\nu+2}\right) =\frac{e^{\nu}(\nu^2+2\nu+2)-e^{\nu}(2\nu+2)}{(\nu^2+2\nu+2)^2} =\frac{e^{\nu}\nu^2}{(\nu^2+2\nu+2)^2}.$$

Therefore,

$$f(t)=\frac13\left[\frac{e^{\nu}}{\nu^2+2\nu+2}\right]_0^{t^3} =\frac13\left(\frac{e^{t^3}}{t^6+2t^3+2}-\frac{1}{2}\right).$$

So,

$$f(t)=\frac{e^{t^3}}{3(t^6+2t^3+2)}-\frac16.$$

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3. Compute $f(1)$

At $t=1$,

$$t^6+2t^3+2=1+2+2=5.$$

Thus,

$$f(1)=\frac{e}{15}-\frac16.$$

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4. Compute $f'(1)$

By the Fundamental Theorem of Calculus,

$$f'(t)=e^{t^3}\frac{t^8}{(t^6+2t^3+2)^2}.$$

Therefore,

$$f'(1)=e\cdot \frac{1}{(1+2+2)^2}=\frac{e}{25}.$$

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5. Add $f(1)+f'(1)$

$$f(1)+f'(1)=\left(\frac{e}{15}-\frac16\right)+\frac{e}{25} = e\left(\frac{1}{15}+\frac{1}{25}\right)-\frac16.$$

Now,

$$\frac{1}{15}+\frac{1}{25}=\frac{5+3}{75}=\frac{8}{75}.$$

So,

$$f(1)+f'(1)=\frac{8e}{75}-\frac16.$$

Comparing with

$$\alpha e-\frac16,$$

we get

$$\alpha=\frac{8}{75}.$$

Hence,

$$150\alpha=150\cdot \frac{8}{75}=16.$$

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6. Comparison with stored answer

Derived answer: $16$

Stored correct answer: $16$

They match.