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

Question

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

Answer 1

$ \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 $\ln{|x+1|} -2\ln{|x-1|} + 8\ln{|x+3|} + C$

Answer 2

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,

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

Answer 3

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

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:

$$= \int\bigg(-\frac{2}{x+1}+\frac{8}{x+3}+\frac{1}{x-1}\bigg) dx\\ \Longrightarrow=\boxed{\color{teal}{\ln|x-1|-2\ln|x+1|+8\ln|x+3|+C}}$$

Answer 4

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.

Answer 5

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

1. the roots of the denominator are all simple
2. the numerator has lower degree than denominator.

you can read off the decomposition directly.

$$\require{cancel} \newcommand{\xxx}[2]{\color{red}{\cancelto{#2}{\color{gray}{#1}}}} \frac{7x^2+1}{(x+1)(x-1)(x+3)}\\ = \frac{\xxx{7(-1)^2+1}{8}}{(x+1)\xxx{(-1-1)(-1+3)}{-4}} + \frac{\xxx{7(1)^2+1}{8}}{(x-1)\xxx{(1+1)(1+3)}{8}} + \frac{\xxx{7(-3)^2+1}{64}}{\xxx{(-3+1)(-3-1)}{8}(x+3)}\\ = \frac{-2}{x+1}+\frac{1}{x-1}+\frac{8}{x+3} $$