So in \(\displaystyle v'=\frac{q}{6L}(L^3x-\frac{3L^2x^2}{2}+Lx^3-\frac{x^4}{4}+C_1\) you mean that when I solve for \(\displaystyle C_1\) it should work out?

I tried this with the following info v'(0)=0 and I got \(\displaystyle C_1=0\).

v'= the slope of the beam deflection curve and v'(0)=0 corresponds to a fixed anchor point.

Assuming I don't have the book answer to use as help, can this problem be done without factoring it out.

These are Mech of Material problems and having to factor will add an extra tedious step, especially on exams.

Thanks again

So I edited my post a few times mostly because I am weird like that but also to clarify a few things:

Technically your answer is right as it was originally because you are only missing a single term that is a constant with respect to x. If a teacher marks that wrong you can contest your answer by saying L^4 is zero when you take its derivative with respect to x so it does not matter if it is hidden in a constant C or not.

Also, the way I wrote it you aren't exactly solving for \(\displaystyle C_{1}\) but rather for another constant that is a combination of \(\displaystyle C_{1}\) and \(\displaystyle C_{2}\). This is a minor nitpicky point but what I hoped to show you was that there was a missing term and it was hidden as an arbitrary constant.

The best way to solve these types of integration problems is to use u-substitution. Since you have \(\displaystyle (L-x)^3\) with respect to \(\displaystyle x\), using u-sub will get rid of the \(\displaystyle x\) and you will end up with a simple varible \(\displaystyle u^3\) with \(\displaystyle -du\) (because if \(\displaystyle u = L - x\) then \(\displaystyle du = -dx\)). Integrating \(\displaystyle -\int u^3du\) is very easy.

Let me know if something doesn't make sense.