Evaluate RS Integral with discontinuous integrator 
Show that
  $$
\int_0^1 \left(x^3+7\right)d \alpha(x) = \frac{29}{2},
\quad \text{where } \alpha(x) = 
\begin{cases}
1+x^3, & 0 < x \le 1\\
0, & x = 0
\end{cases}
$$

I don't know how to handle discontinuous integrators in RS integrals. Please help. I tried to solve it but ended up getting 15/2. Don't know where I went wrong.
Which property can be used here?
I converted the RS integral to R integral but it did not work.
Assuming $f$ and $g$ to be nicely differentiable, I used:
$$\int f(x)dg(x)=\int f(x)g'(x)dx,$$
so
$$
\int_0^1 \left(x^3+7\right)d (1+x^3) = \int_0^1 \left(3x^2)(x^3+7\right)dx= \int_0^1 \left(3x^5+21x^2\right)dx =\frac{15}{2},
$$
However in my textbook the answer is given to be $\frac{29}{2}$.

Your analysis takes care of the continuous part of the integral over $x \in (0,1]$ but there is a discrete point at $x=0$ which will additionally contribute $f(0)\Delta \alpha(0)$. Note that $f(0)=7$ and
$$
\Delta \alpha(0) = \alpha(0^+) - \alpha(0) = \left(1+0^3\right) - 0 = 1.
$$
Hence, the net contribution of the discrete point $x=0$ is
$$
f(0)\Delta \alpha(0) = 7 \cdot 1 = 7,
$$
and the final integral is
$$
f(0)\Delta \alpha(0) + \int_0^1 \left(x^3+7\right) 3x^2 dx
= 7 + \frac{15}{2}
= \frac{29}{2}
$$
as desired.
