<s>[INST] Solve $ \int{\frac{7x^2 + 1}{(x+1)(x-1)(x+3)}}\,dx $ I don’t know how to solve this integral:
$$\int{\frac{7x^2 + 1}{(x+1)(x-1)(x+3)}}\,dx$$
I know this is a rational integral but I don’t know how to write it in a different way
 [/INST] 
$$ \int{\frac{(7x^2 + 1)}{(x+1)(x-1)(x+3)}}dx$$

Partial fractions:
You need to solve the equation: $$ \color{green}7x^2 +\color{red}0\cdot x +\color{blue}1 = A(x-1)(x+3) + B(x+1)(x+3) + C(x+1)(x-1) $$
$$=A(x^2+2x-3)+B(x^2+4x+3)+C(x^2-1)\\=\color{green}{(A+B+C)}x^2+\color{red}{(2A+4B)}x+\color{blue}{(-3A+3B-C)} $$
To find A, B and C you need to solve an simultaneous equation:
$ \color{green}{A + B + C} =  \color{green}7$
$ \color{red}{2A + 4B} = \color{red}0$
$\color{blue}{3A+3B-C} = \color{blue}1$
After solving it you can find out that $ A = -2$, $B=1$ and $ C = 8$
So our integral can be written as:
$$= \int\bigg(-\frac{2}{x+1}+\frac{8}{x+3}+\frac{1}{x-1}\bigg) dx\\ \Longrightarrow=\boxed{\color{teal}{\ln|x-1|-2\ln|x+1|+8\ln|x+3|+C}}$$
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