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

$ \int{\frac{(7x^2 + 1)dx}{(x+1)(x-1)(x+3}} $ = $ \int{\frac{A dx}{(x+1)}} + \int{\frac{B dx}{(x-1)}} + \int{\frac{C dx}{(x+3)}}$
You need to solve the equation:
$ 7x^2 + 1 = A(x-1)(x+3) + B(x+1)(x+3) + C(x+1)(x-1) $
To find A, B and C you need to solve an simultaneous equation:
$A + B + C = 7$
$2A + 4B = 0$
$-3A+3B-C = 1$
After solving it you can find out that $ A = 1$, $B=-2$ and $ C = 8$
So our integral  can be written as:
$ \int{\frac{(7x^2 + 1)dx}{(x+1)(x-1)(x+3}} $ = $ \int{\frac{dx}{(x+1)}} - \int{\frac{2 dx}{(x-1)}} + \int{\frac{8 dx}{(x+3)}}$
Each of this integrals can be solved by substitution, you just need to use $t$ as $x+1$, $x-1$ and $x+3$
For example:
$x+1 = t$, so $dx = dt$
$\int{\frac{dx}{(x+1)}} $= $\int{\frac{dt}{t}}$ = $\ln{|t|} + C $ = $\ln{|x+1|} + C$
When you solve all of these 3 internals you will see that the answer is $\ln{|x+1|} -2\ln{|x-1|} + 8\ln{|x+3|} + C$
