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
 A: 
$$ \int{\frac{(7x^2 + 1)}{(x+1)(x-1)(x+3)}}dx$$

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:
$$= \int\bigg(-\frac{2}{x+1}+\frac{8}{x+3}+\frac{1}{x-1}\bigg) dx\\ \Longrightarrow=\boxed{\color{teal}{\ln|x-1|-2\ln|x+1|+8\ln|x+3|+C}}$$
A: 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,
$$\int \frac{(7x^2 + 1)}{(x+1)(x-1)(x+3)}=\int \left(\frac{-2}{x+1}+\frac{1}{x-1}+\frac{8}{x+3}\right)\,dx$$
$$=-2\log|x+1|+\log|x-1|+8\log|x+3|+C$$
A: Let me elaborate on partial fraction, and give you a way that doesn't involve solving linear systems.
We have
$$\frac{(7x^2 + 1)}{(x+1)(x-1)(x+3)}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{x+3}.$$
Let's multiply both sides by (x+1)(x-1)(x+3); we get
$$7x^2+1 = A(x-1)(x+3)+B(x+1)(x+3)+C(x-1)(x+1).$$
Since this equality must hold for all $x$, let us plug in some nice values of $x$:


*

*$x=-1$ yields $\ \ 7+1 = A(-1-1)(-1+3)$ and so $A=-2$

*$x=+1$ yields $\ \ 7+1 = B(+1+1)(+1+3)$ and so $B=+1$

*$x=-3$ yields $63+1 = C(+3-1)(+3+1)$ and so $C=+8$


With this we finally get
$$\frac{(7x^2 + 1)}{(x+1)(x-1)(x+3)}=-\frac{2}{x+1}+\frac{1}{x-1}+\frac{8}{x+3}.$$
This methods works every time you have (one or more) linear factors. You can solve for as many constants as you have linear factors.
A: 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.
$$\require{cancel}
\newcommand{\xxx}[2]{\color{red}{\cancelto{#2}{\color{gray}{#1}}}}
\frac{7x^2+1}{(x+1)(x-1)(x+3)}\\
= \frac{\xxx{7(-1)^2+1}{8}}{(x+1)\xxx{(-1-1)(-1+3)}{-4}}
+ \frac{\xxx{7(1)^2+1}{8}}{(x-1)\xxx{(1+1)(1+3)}{8}}
+ \frac{\xxx{7(-3)^2+1}{64}}{\xxx{(-3+1)(-3-1)}{8}(x+3)}\\
= \frac{-2}{x+1}+\frac{1}{x-1}+\frac{8}{x+3}
$$
A: $ \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 $\ln{|x+1|} -2\ln{|x-1|} + 8\ln{|x+3|} + C$
