$\int(x+1)(x+2)^4(x+3) \, dx$ is equal to
- A$\frac{(x+1)^2}{2}+\frac{(x+2)^2}{5}+\frac{(x+3)^2}{2}+C$
- B$\frac{(x+2)^7}{7}-\frac{(x+2)^5}{5}+C$
- C$\frac{(x+2)^7}{7}+\frac{(x+2)^5}{5}+C$
- D$\frac{(x+3)^7}{7}-\frac{(x+3)^5}{5}+C$
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