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Below is a problem I did. The answer I got differed from the book by a factor of $2$. An online calculator get the book's answer. I would like to know where I went wrong.

Problem:

Evalaute the following integral: $$ \int \frac{dx}{ \sqrt{9x^2 - 6x + 5} } \,\, dx $$ Answer:

Let $I$ be the integral we are trying to evaluate. \begin{align*} I &= \int \frac{dx } { 3 \sqrt{x^2 - \left( \frac{2}{3} \right) x + \frac{5}{9} } } \\ 3I &= \int \frac{dx } { \sqrt{ \left( x - \frac{1}{3} \right)^2 + \left( \frac{2}{3} \right) ^ 2 } } \\ u &= x - \frac{1}{3} \\ du &= dx \\ 3I &= \int \frac{du } { \sqrt{ u^2 + \left( \frac{2}{3} \right) ^ 2 } } \\ u &= \left( \frac{2}{3} \right) \tan \left( \theta \right) \\ du &= \left( \frac{2}{3} \right) \sec^2 \left( \theta \right) \,\, d\theta \\ % 3I &= \int \frac{ \left( \frac{2}{3} \right) \sec^2 \left( \theta \right) \,\, d\theta } { \sqrt{ \left( \frac{4}{9} \right)\tan^2 \left( \theta \right) + \left( \frac{2}{3} \right) ^ 2 } } \\ 3I &= \int \frac{ \left( \frac{2}{3} \right) \sec^2 \left( \theta \right) \,\, d\theta } { \left( \frac{2}{3} \right)\sqrt{ \tan ^2 \left( \theta \right) + 1 } } \\ 3I &= \int \frac{ \sec^2 \theta \,\, d\theta } { \sec \theta } \\ 3I &= \int \sec \theta \,\, d\theta \end{align*} \begin{align*} 3I &= \ln{| \sec \theta + \tan \theta |} + C_1 \\ 3I &= \ln{| \sqrt{ \tan^2 \theta + 1 } + \tan \theta |} + C_1 \\ 3I &= \ln{\Bigg| \sqrt{ \left( \frac{9}{4} \right) u^2 + 1 } + \left( \frac{3}{2}\right) u \Bigg| } + C_1 \\ 3I &= \ln{\Bigg| \sqrt{ \left( \frac{9}{4} \right) \left( x - \frac{1}{3} \right) ^2 + 1 } + \left( \frac{3}{2}\right) \left( x - \frac{1}{3} \right) \Bigg| } + C_1 \\ I &= \left( \frac{1}{3}\right) \ln{\Bigg| \sqrt{ \left( \frac{9}{4} \right) \left( x - \frac{1}{3} \right) ^2 + 1 } + \left( \frac{3}{2}\right)x - \left( \frac{1}{2} \right) \Bigg| } + C \\ % I &= \left( \frac{1}{3}\right) \ln{\Bigg| \sqrt{ \left( \frac{9}{4} \right) \left( x^2 - \left( \frac{2}{3} \right) x + \frac{1}{9} \right) + 1 } + \left( \frac{3}{2}\right)x - \left( \frac{1}{2} \right) \Bigg| } + C \\ % I &= \left( \frac{1}{3}\right) \ln{\Bigg| \sqrt{ \left( \frac{9}{4}\right) x^2 - \left( \frac{3}{2} \right) x + \frac{5}{4} } + \left( \frac{3}{2}\right)x - \left( \frac{1}{2} \right) \Bigg| } + C \\ I &= \left( \frac{1}{6}\right) \ln{\Bigg| \sqrt{ 9x^2 - 6x + 5 } + 3x - 1 \Bigg| } + C \\ \end{align*} The book's answer is: $$ \left( \frac{1}{3} \right) \ln{| \sqrt{9x^2 - 6x + 5} + 3x - 1 |} + C $$

Based upon comments from the group, I updated the last step. My answer is now: $$ I = \left( \frac{1}{3}\right) \ln{\Bigg| \frac{ \sqrt{ 9x^2 - 6x + 5 } + 3x - 1 }{2} \Bigg| } + C $$ However, this answer is still wrong. I would like to know where I went wrong.

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    $\begingroup$ Why did you half the factor at the front in the last step? It's not necessary to do so. $\endgroup$ Commented Jul 19, 2020 at 12:47
  • $\begingroup$ @PeterForeman I half the factor in the last step because I took out a $\frac{1}{2}$ from the square root sign. However, looking at it again that was wrong. $\endgroup$ Commented Jul 19, 2020 at 12:50
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    $\begingroup$ What you should have is something like$$\frac13\ln{(z/2)}+C_1=\frac13\ln{(z)}+C_2$$where $z=\left|\sqrt{9x^2-6x+5}+3x-1\right|$ and $C_2=C_1-\frac13\ln{(2)}$. $\endgroup$ Commented Jul 19, 2020 at 12:55
  • $\begingroup$ @PeterForeman I have updated my answer but it is still wrong. I believe my error is not in the last two steps. $\endgroup$ Commented Jul 19, 2020 at 13:41
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    $\begingroup$ Your error is in the last two steps as I said above. The second to last line is equivalent to the correct answer. $\endgroup$ Commented Jul 19, 2020 at 16:07

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This integral fits into the standard form $$\int \frac{dx}{\sqrt{a^2+x^2}} = \ln|x + \sqrt{x^2 + a^2}| + c$$ Your integral can be rewritten as: $$I = \int\frac{dx}{\sqrt{(3x-1)^2 + 4}}$$ Let $3x-1 = t \implies 3dx = dt$ $$I = \frac{1}{3}\int\frac{dt}{\sqrt{t^2 + 4}}$$ Can you take it from here?

EDIT: As somebody pointed out in the comments, you got the correct answer, but made a mistake in the final few steps. $$\frac{1}{3}\ln\left| \frac{\sqrt{9x^2 - 6x + 5} + 3x-1}{2}\right| \ne \frac{1}{6}\ln\left|\sqrt{9x^2 - 6x + 5} + 3x-1\right|$$

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  • $\begingroup$ I have updated my answer but it is still wrong. I would like to be able to do this problem without using the standard integral you cited. I am hoping that you can tell me where I went wrong. It is true that you found a mistake that I made but it is not the only mistake I made in this problem. $\endgroup$ Commented Jul 19, 2020 at 13:45
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    $\begingroup$ Your answer is right. You can rewrite it as $$\frac{1}{3}\ln\left|\sqrt{9x^2 - 6x + 5} + 3x-1\right| - \frac{1}{3}\ln(2) + C = \frac{1}{3}\ln\left|\sqrt{9x^2 - 6x + 5} + 3x-1\right| + C'$$, which is the correct answer. $\endgroup$ Commented Jul 19, 2020 at 13:58

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