How to find $\int_0^1 x^4(1-x)^5dx$ quickly? This question came in the Rajshahi University admission exam 2018-19
Q) $\int_0^1 x^4(1-x)^5dx$=?
(a) $\frac{1}{1260}$
(b) $\frac{1}{280}$
(c)$\frac{1}{315}$
(d) None
This is a big integral (click on show steps):
$$\left[-\dfrac{\left(x-1\right)^6\left(126x^4+56x^3+21x^2+6x+1\right)}{1260}\right]_0^1=\frac{1}{1260}$$
It takes a lot of time to compute. How can I compute this quickly (30 seconds) using a shortcut?

I wouldn't say this method helps you in solving it in 30 seconds, but I think it can help you in simplifying the calculations so that the integral can be computed faster
First consider by usage of the Binomial Theorem
$$(1-x)^5=\sum_{k=0}^{5}\binom{5}{k}(1)^k(-x)^{5-k}=\sum_{k=0}^{5}\binom{5}{k}(-1)^{5-k}(x)^{5-k}$$
from which you can get by multiplying by $x^4$
$$x^4(1-x)^5=\sum_{k=0}^{5}\binom{5}{k}(-1)^{5-k}(x)^{9-k}$$
applying the integral to the sum
$$\int_{0}^{1}x^4(1-x)^5\cdot dx=\sum_{k=0}^{5}\binom{5}{k}(-1)^{5-k}\bigg(\frac{x^{10-k}}{10-k} \bigg)_{0}^{1}$$
which will give
$$\int_{0}^{1}x^4(1-x)^5\cdot dx=\sum_{k=0}^{5}\binom{5}{k}(-1)^{5-k}\bigg(\frac{1}{10-k} \bigg)$$
Expanding the sum gives
$$-\binom{5}{0}\bigg(\frac{1}{10}\bigg)+\binom{5}{1}\bigg(\frac{1}{9}\bigg)-\binom{5}{2}\bigg(\frac{1}{8} \bigg)+\binom{5}{3}\bigg(\frac{1}{7}\bigg)-\binom{5}{4}\bigg(\frac{1}{6} \bigg)+\binom{5}{5}\bigg(\frac{1}{5} \bigg)$$
simplifying will give
$$\frac{-1}{10} +\frac{5}{9} -\frac{10}{8} +\frac{10}{7} -\frac{5}{6} +\frac{1}{5}=\frac{1}{1260}$$
