<s>[INST] How can I integrate $\int {dx \over \sqrt{3^2+x^2}} $ using Trigonometric Substitution? $$\int {dx \over \sqrt{9+x^2}} = \int {dx \over \sqrt{3^2+x^2}} $$
$$ x =3\tan\theta$$
$$dx = 3\sec^2\theta$$
$$\int {3\sec^2\theta \over \sqrt{3^2 + 3^2\tan^2\theta}} d\theta$$
$$\int {3\sec^2\theta \over \sqrt{3^2(1+\tan^2\theta)}} d\theta$$
$$\int {3\sec^2\theta \over 3\sec\theta} d\theta = \int \sec\theta$$
$$\ln|\sec\theta + \tan\theta| + C $$
$$\ln\left({\sqrt{9+x^2} \over 3} + {x \over 3}\right)$$
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?
 [/INST] Hint: $$\log (t\times C) = \log t + \log C$$
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