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$$\int {dx \over \sqrt{9+x^2}} = \int {dx \over \sqrt{3^2+x^2}} $$ $$ x =3\tan\theta$$ $$dx = 3\sec^2\theta$$

$$\int {3\sec^2\theta \over \sqrt{3^2 + 3^2\tan^2\theta}} d\theta$$ $$\int {3\sec^2\theta \over \sqrt{3^2(1+\tan^2\theta)}} d\theta$$ $$\int {3\sec^2\theta \over 3\sec\theta} d\theta = \int \sec\theta$$ $$\ln|\sec\theta + \tan\theta| + C $$ $$\ln\left({\sqrt{9+x^2} \over 3} + {x \over 3}\right)$$

My book says the answer should just be:

$$\ln\left({\sqrt{9+x^2}} + {x}\right) $$

I'm wondering where I went wrong with this?

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    $\begingroup$ To obtain $\sin x$, type \sin x in math mode. Similarly, type \cos x, \tan x, \sec x, \csc x, \cot x, \ln x, \log x in math mode to obtain $\cos x$, $\tan x$, $\sec x$, $\csc x$, $\cot x$, $\ln x$, and $\log x$, respectively. $\endgroup$
    – N. F. Taussig
    May 1, 2015 at 10:36

3 Answers 3

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The two results are equivalent, up to choice of constant.

\begin{align} \ln\left({\sqrt{9+x^2}+x \over 3}\right)+C &= \ln\left(\sqrt{9+x^2} + x\right) - \ln(3) + C \\ &= \ln\left(\sqrt{9+x^2} + x\right) + C' \end{align}

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  • $\begingroup$ Ok I see that, thanks $\endgroup$
    – nukenine
    Apr 30, 2015 at 20:48
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Hint: $$\log (t\times C) = \log t + \log C$$

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It should be a printing mistake. Your answer is correct

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  • $\begingroup$ As you can see from the other answers, though the OP's answer is correct, so is the book's. There was no printing mistake. $\endgroup$
    – Rory Daulton
    Apr 30, 2015 at 21:28

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