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

Hint:
Partial Fractions yields:
$$\frac{(7x^2 + 1)}{(x+1)(x-1)(x+3)}=\frac{-2}{x+1}+\frac{1}{x-1}+\frac{8}{x+3}$$
Therefore,
$$\int \frac{(7x^2 + 1)}{(x+1)(x-1)(x+3)}=\int \left(\frac{-2}{x+1}+\frac{1}{x-1}+\frac{8}{x+3}\right)\,dx$$
$$=-2\log|x+1|+\log|x-1|+8\log|x+3|+C$$
