Evaluation of $\int x^{26}(x-1)^{17}(5x-3)dx$

Note that \begin{align} \frac{d}{dx}\left[\color{blue}{\frac{1}{9}x^{27}(x-1)^{18}}\right]&=2x^{27}(x-1)^{17}+3x^{26}(x-1)^{18}\\ &=x^{26}(x-1)^{17}(2x+3(x-1))\\ &=x^{26}(x-1)^{17}(5x-3). \end{align} Intuition: given the form of the integrand, I played around with $Cx^{27}(x-1)^{18}$ and found $C=\frac{1}{9}$ worked.

You have this: $$\left(\begin{array}{l} \text{first-degree} \\ \text{polynomial} \end{array} \right)^\text{large exponent} \cdot \left(\begin{array}{l} \text{first-degree} \\ \text{polynomial} \end{array} \right)^\text{large exponent} \cdot \left(\begin{array}{l} \text{first-degree} \\ \text{polynomial} \end{array} \right)$$

Perhaps the only simple way to do this is to recall what you see above is what you get when you evaluate $$ \frac d {dx } \left( \left(\begin{array}{l} \text{first-degree} \\ \text{polynomial} \end{array} \right)^\text{large exponent} \cdot \left(\begin{array}{l} \text{first-degree} \\ \text{polynomial} \end{array} \right)^\text{large exponent} \right) $$ For example, \begin{align} & \frac d {dx} (4x+19)^{42} (2x-27)^{50} \\[10pt] = {} & \underbrace{\overbrace{42(4x+19)^{41}\cdot4\cdot (2x-27)^{50}} {}+{} \overbrace{(4x+19)^{42} 50(2x-27)^{49}\cdot 2}}_\text{product rule} \\[10pt] = {} & \Big( \underbrace{(4x+19)^{41} (2x-27)^{49}}_\text{the common factor} \Big) \cdot \Big( \text{whatever is left (a sum of two terms, admitting simplification)} \Big) \\[10pt] = {} & \left(\begin{array}{l} \text{first-degree} \\ \text{polynomial} \end{array} \right)^\text{large exponent} \cdot \left(\begin{array}{l} \text{first-degree} \\ \text{polynomial} \end{array} \right)^\text{large exponent} \cdot \left( \underbrace{ \begin{array}{l} \text{first-degree} \\ \text{polynomial} \end{array} }_\text{This is “whatever is left.''} \right) \end{align}