Integration by Substitution I have the following problem: $\int (x+2)(2x-3)^6dx$
What I did was allow $u = 2x-3$
$dx= \frac{1}{2}du$
Since I am integrating w.r.t u I decided to let $x=\frac{u+3}{2}$
my new equation is $\int ( \frac{u+3}{2}+2)(u^6)\frac{1}{2}du$
which I then simplify to $\int ( \frac{u+7}{4})(u^6)du$
I think I did something wrong because my following steps aren't giving the correct answer 

You did all the work here (the hard part).... To finish it off, note that
$$\begin{align} \int \left( \frac{u+7}{4}\right)(u^6)\,du &= \frac 14\int(u+7)(u^6)\,du \\ \\ &= \frac 14\int (u^7 + 7u^6)\,du \\ \\ &= \frac 14\left(\frac{u^8}{8} + u^7\right) + C \\ \\ &= \frac 14 u^7\left(\frac u8 + 1\right)\end{align}$$
All that's left for you is to back substitute by replacing $u$ with $2x-3$.
