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How to continue?

$$\int_{1}^3{\frac{dx}{x^2-6x+8}} =$$ $$ =\left[ \frac12 \ln\left|\frac{x-4}{x-2}\right| \right]_{1}^3 =$$ $$= \frac12 \ln\left|\frac{-1}{1} \right| - \frac12 \ln\left|\frac{-3}{-1} \right|$$

How to solve integrals like this, when it is not define?

DavidM
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  • You cannot perform the integration, because the integral does not converge in that range. – Enrico M. Jan 20 '16 at 20:48
  • How should I test it if it converge in that range? – DavidM Jan 20 '16 at 20:53
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    Solving $x^2 - 6x + 8$ gives you two solutions:

    $$x_{1; 2} = \frac{6\pm \sqrt{36 - 32}}{2} = \frac{6\pm 2}{2}$$

    So $x_1 = 4$ and $x_2 = 2$

    the second value, $x=2$ belongs to the extrema of integration, so there is a pole along the integration path, which makes the integral to diverge.

     – Enrico M. Jan 20 '16 at 21:05

3 Answers3

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$\ln(1)$ is defined. It is $0$. You may be thinking of $\ln(0)$, which is not defined.

Mankind
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Unfortunately, the function cannot be integrated because it is too large near $x=2$. Plot the function, and you will see.

Empy2
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  • Yes I saw it, but how is it possible know it without plot the function using wolframalpha? – DavidM Jan 20 '16 at 20:37
  • Factor the polynomial in the denominator. It is $(x-2)(x-4)$, so the denominator goes to zero at $x=2$. – Empy2 Jan 20 '16 at 20:39
  • Thanks, it means that it cannot be integrated for numbers witch give me 0 in the denominator in original function? – DavidM Jan 20 '16 at 20:46
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About your last question: "it means that it cannot be integrated for numbers witch give me 0 in the denominator in original function?", the answer is: Depends. The reason because this particular function is not integrable in [1,3] is because it's not bounded,which you can see with Wolfram if you wish, but there's another way of knowing it: Since the denominator is a polynomial, it is continuous. Besides, it becomes $0$ when $x=2$. This means $\lim_{ x \rightarrow 2} x^2-6x+8=0$. So the denominator approaches zero as much as you wish by making x near 2, and so $r(x)=\dfrac{1}{x^2-6x+8}$ goes to infinity, meaning r(x) is not bounded near 2. However if we consider the function: $f:[1,3]\rightarrow \mathbb{R}$, $f(x)=\dfrac{1}{1+x^2}$ if $x\not =2$ and $f(2)=0$, then it is true that the denominator gives $0$ in $x=2$, but it's clearly integrable on $[1,3]$.

Diego
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