calculus - How to do this integral without the inverse hyperbolic tan

We have the integral $\displaystyle\int \dfrac{1}{x^2+4x+3} dx$. I figured out that with a substitution you can rewrite this as $ - \displaystyle \int \dfrac{1}{1-u^2} du$, which is $-\tanh^{-1}(u)$. However, I wonder if there's another way to do this integral without having to use the (in my opinion) obscure inverse hyperbolic trig functions.

Answer

You can apply parital fractions to this problem and then simply integrate fractions by logarithms.

$$\frac{1}{x^2+4x+3}=\frac{1}{2(x+1)}-\frac{1}{2(x+3)}$$