<s>[INST] Where does the integration end? I'm new to integrals. I'm solving $$ \int \frac{1}{2x^2+6}$$ but I get a wrong answer: $$ \frac{1}{6}\arctan\frac{x}{\sqrt3}$$
The correct answer should be: $$ \frac{\sqrt3}{6}\arctan\frac{x}{\sqrt3}$$
Here is my full try:
$$ \int \frac{1}{6(\frac{2x^2}{6}+1)} = \int \frac{1}{6(1+(\frac{x}{\sqrt3})^2)} = \frac{1}{6}\arctan\frac{x}{\sqrt3}$$
Can you correct me and give me some source to learn from?
Thanks in advance!
 [/INST] $$\int \frac{1}{2x^2+6}dx = \frac{1}{2}\int \frac{1}{x^2+3}dx$$
$$x = \sqrt{3}\tan{\theta}\Rightarrow dx = \sqrt{3}\sec^2{\theta}d\theta$$
Plugging our substitution back into the integral yields
$$\frac{\sqrt{3}}{2}\int \frac{\sec^2{\theta}}{3\tan^2{\theta}+3}d\theta = \frac{\sqrt{3}}{6}\int \frac{\sec^2{\theta}}{\sec^2{\theta}}d\theta$$
So we are now left with
$$\frac{\sqrt3}{6}\theta +c$$
Since this is an indefinite integral, we have to write our answer in terms of x. Looking back at our substitution and rearranging for theta, we get to our final answer:
$$\frac{\sqrt3}{6}\tan^{-1}(\frac{x}{\sqrt{3}})+c$$
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