# Math Help - integration by partial fractions

1. ## integration by partial fractions

I am having some trouble with $\int\frac{x+1}{x^2+4x+3}~dx$. I can break it down to $\frac{x+1}{x^2+4x+3}=\frac{A}{x+1}+\frac{B}{x+3}$. 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, $\int\frac{x+1}{x^2+4x+3}~dx=\int\frac{x+1}{(x+1)(x +3)}~dx=\int\frac1{x+3}~dx$. And therefore is this what I am really integrating?

Just making sure my logic is correct.

2. Originally Posted by Possible actuary
I am having some trouble with $\int\frac{x+1}{x^2+4x+3}~dx$. I can break it down to $\frac{x+1}{x^2+4x+3}=\frac{A}{x+1}+\frac{B}{x+3}$. 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, $\int\frac{x+1}{x^2+4x+3}~dx=\int\frac{x+1}{(x+1)(x +3)}~dx=\int\frac1{x+3}~dx$. And therefore is this what I am really integrating?

Just making sure my logic is correct.
you are correct. the last factorization method is what i would use. don't worry that things don't look exactly the same algebraically. what we actually did was take 1/(x + 3) and multiply it by 1, namely by (x + 1)/(x + 1).

if you plot a graph of (x + 1)/(x^2 + 4x + 3), you would get exactly the same graph as the graph for 1/(x + 3)

3. Originally Posted by Jhevon
you are correct. the last factorization method is what i would use. don't worry that things don't look exactly the same algebraically. what we actually did was take 1/(x + 3) and multiply it by 1, namely by (x + 1)/(x + 1).

if you plot a graph of (x + 1)/(x^2 + 4x + 3), you would get exactly the same graph as the graph for 1/(x + 3)
But make sure your integration region doesn't include x = -1! Because $\frac{x +1}{x^2 + 4x + 3}$ doesn't exist at x = -1. $\frac{1}{x + 3} \neq \frac{x +1}{x^2 + 4x + 3}$. The domains of the two expressions are different.

-Dan