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?

• Technical nitpick: Don't forget the $dx$, and done forget the ${}+C$. They may both seem somewhat unnecessary when you first start out with integration, but they are actually both pretty important. Commented Aug 16, 2020 at 19:10
• There's no such thing as $$\int\frac1{2x^2+6}.$$ There is such a thing as $$\int\frac{dx}{2x^2+6}.$$ That $dx$ is important. You need to account for it when doing the substitution. Commented Aug 16, 2020 at 19:10
• The $dx$ is certainly important when you do a substitution since $dx$ also changes. In integration theory it's however common to write just $\int f.$ But then the dummy variable is not mentioned at all. Commented Aug 16, 2020 at 22:17

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.

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,

$$\bbox[5px,border:2px solid red]{\frac{\sqrt3}{6}\arctan\frac{x}{\sqrt3}}$$

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 $$\sqrt3\arctan\left(\frac x{\sqrt3}\right)$$.

• Excuse me very much but is it $\sqrt3/6$ or $\sqrt3$ the coefficient of $\arctan$ (see the previous answer). Commented Aug 16, 2020 at 19:14
• The other answer says that$$\int\frac1{6\left(1+\left(\frac x{\sqrt3}\right)^2\right)}\,\mathrm dx=\frac{\sqrt3}6\arctan\left(\frac x{\sqrt3}\right).$$My answer says that$$\int\frac1{1+\left(\frac x{\sqrt3}\right)^2}\,\mathrm dx=\sqrt3\arctan\left(\frac x{\sqrt3}\right).$$It's the same thing. Commented Aug 16, 2020 at 19:20
• Excuse me very much: I have not seen the $6$ that there is not into your last integral of the comment. Commented Aug 16, 2020 at 19:23
• @Sebastiano No problem. It's a natural oversight. Commented Aug 16, 2020 at 20:18
• No, no 😩😩 It is my head that it's my head that's burnt out. Also the temperature in Sicily it is similar to Sahara desert. Commented Aug 16, 2020 at 20:44

$$\int \frac{1}{2x^2+6}dx = \frac{1}{2}\int \frac{1}{x^2+3}dx$$

$$x = \sqrt{3}\tan{\theta}\Rightarrow dx = \sqrt{3}\sec^2{\theta}d\theta$$

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:

$$\frac{\sqrt3}{6}\tan^{-1}(\frac{x}{\sqrt{3}})+c$$

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}