Integration of $\int_1^2 x(2x-3)^4 \, dx$ by substitution An exam question: $$\int_1^2 x(2x-3)^4 \, dx\\
U = 2x - 3$$
I have rearranged to get $dx = dU/2$. So I am now at
$\int xU^4 \, dU$
I am not quite sure what to do with the $x$ as it is not cancelled out as I thought it would. Thanks for any help.
 A: You need to substitute every instance of $x$.
You are then at
$$
\int \frac{U+3}2 U^4 \frac{dU}2$$
and then use the formula
$$
\int t^m dt = \frac {t^{m+1}}{m+1}
$$
Can you take it from here?
A: Using capital $U$ and lower-case $u$ interchangeably gets perceived by mathematicians the way others perceive a spelling error, and moreover in many contexts the corresponding capital and lower-case letters refer to two different things in the same problem.  You've probably seen this in trigonometry where $a$ is the length of the side of a triangle and $A$ is the measure of the angle that is opposite that side.  In this sort of problem lower case is somewhat customary and I use it below:
\begin{align}
u & = 2x-3 \\[6pt]
du & = 2\,dx \\[6pt]
x & = \frac{u-3} 2 \\[6pt]
\int x(2x-3)^4\,dx & = \int\frac{u-3} 2 u^4 \left(\frac{du} 2\right) = \frac 1 4\int(u^5 - 3u^4)\,du = \cdots\cdots
\end{align}
A: In the same spirit as Mookid's answer, consider a more general case $$I=\int (A+Bx)(C+Dx)^n\,dx$$ Change variable $C+Dx=u$, $x=\frac{u-C}{D}$, $dx=\frac{du}{D}$ $$I=\int\big(A+\frac{B (u-C)}{D}\big)u^n \,du=\int\big(A-\frac{B C}{D}+\frac{B u}{D})u^n\,du$$ $$I=\big(A-\frac{B C}{D}\big)\int u^n \, du+\frac{B }{D}\int u^{n+1} \, du$$ $$I=\big(A-\frac{B C}{D}\big)\frac{u^{n+1}}{n+1}+\frac{B }{D}\frac{u^{n+2}}{n+2}$$ Now, as jnh commented, $n$ can be as large as you want, rational, irrational, it does not make any difference.
A: HINT:$$x(2x-3)^4=16\,{x}^{5}-96\,{x}^{4}+216\,{x}^{3}-216\,{x}^{2}+81\,x$$
