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!
 A: You are correct all the way up until (and including) the step:
$${=\int \frac{1}{6\left(1 + \left(\frac{x}{\sqrt{3}}\right)^2\right)}dx}$$
You are incorrectly applying the fact that
$${\int \frac{1}{1+x^2}dx=\arctan(x)+c}$$
Notice it must be ${1+x^2}$ - not ${1+ax^2}$. Instead, you should then do the substitution ${u=\frac{x}{\sqrt{3}}}$ to get
$${=\frac{\sqrt{3}}{6}\int\frac{1}{1+u^2}du=\frac{\sqrt{3}}{6}\arctan(u)+c=\frac{\sqrt{3}}{6}\arctan\left(\frac{x}{\sqrt{3}}\right)+c}$$
As required.
A: Given, $$\int \frac{1}{2x^2+6}$$
We know that,$$\int{\frac{1}{a^2+u^2}}dx = \frac{1}{a}\tan^{-1}(\frac{u}{a})+c$$
So,
$$\int \frac{1}{6(\frac{2x^2}{6}+1)}dx $$
$$= \int \frac{1}{6(1+(\frac{x}{\sqrt3})^2)}dx$$
Here,$a=1$ and $u=\frac{x}{\sqrt3}$ and $du=\frac{dx}{\sqrt3}$,
i.e, $dx={\sqrt3}du$
So our desired answer is,
$$\bbox[5px,border:2px solid red]{\frac{\sqrt3}{6}\arctan\frac{x}{\sqrt3}}$$
A: Your problem lies in the final equality. If $F(x)$ is a primitive of $f(x)$, and if $c\ne0$, then a primitive of $f(cx)$ will be $\frac1cF(cx)$. So, since $\arctan(x)$ is a primitive of $\frac1{1+x^2}$, a primitive of $\frac1{1+(x/\sqrt3)^2}$ will be $\sqrt3\arctan\left(\frac x{\sqrt3}\right)$.
A: $$\int \frac{1}{2x^2+6}dx = \frac{1}{2}\int \frac{1}{x^2+3}dx$$
$$x = \sqrt{3}\tan{\theta}\Rightarrow dx = \sqrt{3}\sec^2{\theta}d\theta$$
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:
$$\frac{\sqrt3}{6}\tan^{-1}(\frac{x}{\sqrt{3}})+c$$
A: Substitute $x=\sqrt3\,u$
$$
\begin{align}
\int\frac{\mathrm{d}x}{2x^2+6}
&=\frac{\sqrt3}6\int\frac{\mathrm{d}u}{u^2+1}\\
&=\frac{\sqrt3}6\arctan(u)+C\\
&=\frac{\sqrt3}6\arctan\left(\frac{x}{\sqrt3}\right)+C\\
\end{align}
$$
