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

I did not understand what substution i have used so that it can simplify,
I have seems it is a derivative of some function.
Help me, Thanks 
 A: 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.
A: 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}
A: Sorry for the informal writing am rushing out.
Hope it helps!
$$
(x - 1)^{17} = \sum \left(
    \begin{array}{c}
      17 \\
      r
    \end{array}
  \right) x^{17-r}(-1)^r
\\
x^{26}(x - 1)^{17} = \sum \left(
    \begin{array}{c}
      17 \\
      r
    \end{array}
  \right) x^{43-r}(-1)^r
\\
 x^{26}(x - 1)^{17}(5x−3) = 5\sum \left(
    \begin{array}{c}
      17 \\
      r
    \end{array}
  \right) x^{45-r}(-1)^r\space- 3\sum\left(
    \begin{array}{c}
      17 \\
      r
    \end{array}
    \right) x^{44-r}(-1)^r
\\
=5x^{45} - \left(\sum\left[
                     5\left( \begin{array}{c}
      17 \\
      r+1
    \end{array}\right) - 3\left( \begin{array}{c}
      17 \\
      r
    \end{array}\right)
               \right]x^{44-r}(-1)^r\right)
+3x^{27}
\\
\int x^{26}(x - 1)^{17}(5x−3) = 5\int x^{45} - \left(\sum\left[
                     5\left( \begin{array}{c}
      17 \\
      r+1
    \end{array}\right) - 3\left( \begin{array}{c}
      17 \\
      r
    \end{array}\right)
               \right]\int x^{44-r}(-1)^r\right)
\\+3\int x^{27} + c
$$
