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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

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5 Answers 5

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

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    $\begingroup$ How is this getting upvoted when the answer is incorrect $\endgroup$
    – user174622
    Commented Jan 1, 2016 at 18:59
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    $\begingroup$ Your partial fraction decomposition is wrong. Why not use the cover-up rule, BTW? $\endgroup$ Commented Jan 1, 2016 at 19:40
  • $\begingroup$ What's the cover-up rule, @DavidQuinn? $\endgroup$ Commented Jan 1, 2016 at 20:36
  • $\begingroup$ @YoTengoUnLCD have a look at this qc.edu.hk/math/Advanced%20Level/Cover%20up%20rule.htm or this meikleriggs.org.uk/CUR/index.php $\endgroup$ Commented Jan 1, 2016 at 21:07
  • $\begingroup$ See achille hui's answer - this is the cover up rule -@YoTengoUnLCD $\endgroup$ Commented Jan 1, 2016 at 21:12
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$$ \int{\frac{(7x^2 + 1)}{(x+1)(x-1)(x+3)}}dx$$

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:

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

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    $\begingroup$ +1 for the correct solution, but it is a bit more than a hint, wouldn't you say? $\endgroup$ Commented Jan 1, 2016 at 19:43
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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,

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

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  • $\begingroup$ Modulus signs, perhaps? $\endgroup$ Commented Jan 1, 2016 at 19:42
  • $\begingroup$ @DavidQuinn Forgot them. Fixed now. $\endgroup$
    – user174622
    Commented Jan 1, 2016 at 19:43
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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.

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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.

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

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