How to evaluate $ \int_{0}^2 x^{26} (x-1)^{17} (5x-3)dx ~~ ?$

Let $$ I = \int_{0}^2 x^{26} (x-1)^{17} (5x-3)dx = \int_{0}^2 x^{26} (x-1)^{17} (2x+3(x-1) )dx = I_1 + I_2 $$ Let us integrate $I_1$ by parts, taking $x^{27} $ as first function, therefore, $$ I_1 = \left . \frac{1}{9} x^{27} (x-1)^{18} \right| _{0}^{2}- I_2 ~ = \frac{2^{27}}{9} -I_2 $$ Hence, $$ I = \frac{2^{27}}{9}~~ .$$

There is a trick, the numbers have been "arranged":

$$(x^{27}(x-1)^{18})'=27x^{26}(x-1)^{17}+18x^{27}(x-1)^{16}=(27(x-1)+18x)x^{26}(x-1)^{17}\\=9(5x-3)x^{26}(x-1)^{17}.$$