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

As every one suggests, do the partial fraction decomposition. However, since 


*

*the roots of the denominator are all simple

*the numerator has lower degree than denominator.


you can read off the decomposition directly.
$$\require{cancel}
\newcommand{\xxx}[2]{\color{red}{\cancelto{#2}{\color{gray}{#1}}}}
\frac{7x^2+1}{(x+1)(x-1)(x+3)}\\
= \frac{\xxx{7(-1)^2+1}{8}}{(x+1)\xxx{(-1-1)(-1+3)}{-4}}
+ \frac{\xxx{7(1)^2+1}{8}}{(x-1)\xxx{(1+1)(1+3)}{8}}
+ \frac{\xxx{7(-3)^2+1}{64}}{\xxx{(-3+1)(-3-1)}{8}(x+3)}\\
= \frac{-2}{x+1}+\frac{1}{x-1}+\frac{8}{x+3}
$$
