What is $\int_0^{1}\frac{x^{300}}{1+x^2+x^3}dx$ upto $2$ decimal places?

In an examination, I was asked to calculate $$\int_0^{1}\frac{x^{300}}{1+x^2+x^3}dx$$. Options were gives as

a - 0.00
b - 0.02
c - 0.10
d - 0.33
e - 1.00

Just look at the questions I felt the integral $$I \geq \sum_0^1{}\frac{x^{300}}{1+x^2+x^3} = 0.33$$. I felt, since numerator is very small as compared to denominator therefore, for value $$\epsilon<1$$, $$1.00$$ isn't possible. So, I chose option d. But I am not sure whether its correct or not as I didn't follow standard procedure.

what is the correct answer and How can it be solved using a standard procedure?

Since $$1+x^2+x^3 \ge 1$$ for $$x \ge 0$$, we have $$\int_0^{1}\frac{x^{300}}{1+x^2+x^3} \, dx \le \int_0^{1}x^{300} \, dx =\frac{1}{301} <\frac{1}{300} =0.00333\cdots$$ This is enough to answer the question.
Since $$1+x^2+x^3 \le 3$$ for $$x \ge 0$$, we have $$\int_0^{1}\frac{x^{300}}{1+x^2+x^3} \, dx \ge \int_0^{1} \frac{1}{3} x^{300} \, dx =\frac{1}{903}$$ Thus $$0.001107 < \frac{1}{903} \le \int_0^{1}\frac{x^{300}}{1+x^2+x^3} \, dx \le \frac{1}{301} < 0.003323$$
• The integral is approximately $0.00111357$ according to WA. – lhf Dec 19 '18 at 11:27
Since $$\frac{x^{300}}{1+x^2+x^3}$$ has a zero of order $$300$$ at the origin, most of the mass of the integral $$\int_{0}^{1}\frac{x^{300}}{1+x^2+x^3}\,dx$$ comes from a neighbourhood of the right endpoint of the integration range. Obviously $$\int_{0}^{1}\frac{x^{300}}{3}\,dx=\frac{1}{903}$$, and
$$I-\frac{1}{903}=\int_{0}^{1}x^{300}\left(\frac{1}{1+x^2+x^3}-\frac{1}{3}\right)\,dx =\int_{0}^{1}x^{300}(1-x)\frac{2+2x+x^2}{3(1+x^2+x^3)}\,dx$$ is bounded by $$\int_{0}^{1}x^{300}(1-x)\,dx = \frac{1}{301\cdot 302}$$ hence the wanted integral is $$\color{green}{0.0011}$$(unknown digits).