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**Possible actuary** I am having some trouble with $\displaystyle \int\frac{x+1}{x^2+4x+3}~dx$. I can break it down to $\displaystyle \frac{x+1}{x^2+4x+3}=\frac{A}{x+1}+\frac{B}{x+3}$. Then let x=-3, and get B=1. Then let x=-1, and get A=0. A=0 will not work because it will then integrate to ln lx+3l + C and that will not differentiate to what I had originally. However, $\displaystyle \int\frac{x+1}{x^2+4x+3}~dx=\int\frac{x+1}{(x+1)(x +3)}~dx=\int\frac1{x+3}~dx$. And therefore is this what I am really integrating?

Just making sure my logic is correct.