Is integration by parts the best method for $\int_0^1 x^3(1-x)^6 dx$?

The beta function is the best idea.

$$\beta(a, b) = \int_{0}^{1} x^{a-1}(1-x)^{b-1} dx$$

For here $I$, let $a = 4, b = 7$

Using:

$$\beta(a, b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$

$$\beta(4, 7) = \frac{\Gamma(4)\Gamma(7)}{\Gamma(11)} = \frac{3!6!}{10!}$$

$$= \frac{3\cdot2\cdot1}{10\cdot9\cdot8\cdot7} = \frac{1}{840}$$

This way might be faster, but it ultimately depends on your personal preference.

Let $u = 1-x$. Then the integral becomes:

$$\int (u-1)^3u^6 \ \text{d}u$$

And that expansion is much easier to work with than the original!

By symmetry,

\begin{align} \int_0^1 x^3(1-x)^6\,\mathrm{d}x &= -\int_1^0x^6(1-x)^3\mathrm{d}x\\ &=-\int_0^1 (x-1)^3x^6\,\mathrm{d}x \\ &=-\int_0^1 x^9-3x^8+3x^7-x^6 \,\mathrm{d}x\\ &=-\left(\frac{1}{10}-\frac{1}{3}+\frac{3}{8}-\frac{1}{7}\right)+0\\\\ &=-\frac{84-280+315-120}{840}\\ &=\boxed{\displaystyle\frac{1}{840}} \end{align}