How to Integrate [1/(x^2 + 3)] dx?

[askor]

What is $\int \frac{1}{x^2 + 3} \ dx$?

This is my attempt:

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

$x^2 + 3 = (\sqrt{3} \tan \theta)^2 + 3$
$= 3 \tan^2 \theta + 3$
$= 3 (\tan^2 \theta + 1)$
$= 3 \sec^2\theta$

$\int \frac{1}{x^2 + 3} \ dx = \int \frac{1}{3 \sec^2\theta} (\sqrt{3} \sec^2 \theta \ d\theta)$
$= \frac{\sqrt{3}}{3} \int d\theta$
$= \frac{\sqrt{3}}{3} \theta + C$

Is there any next steps?

Is this correct?

Thanks

 

[PeroK]

You can always check an integral for yourself by differentiating your answer to see whether you get the original integrand.

 

[askor]

I know, but I am still don't understand what should I do with the theta.

 

[PeroK]

I know, but I am still don't understand what should I do with the theta.

You have $x = \sqrt 3 \tan \theta$. That means that $\theta = \tan^{-1}\big (\dfrac x {\sqrt 3} \big )$

 

[askor]

Thank you very much, now I understand. So the final result is $\frac{\sqrt{3}}{3} \tan^{-1} \left( \frac{\sqrt{3}x}{3} \right) + C$

 

[PeroK]

Thank you very much, now I understand. So the final result is $\frac{\sqrt{3}}{3} \tan^{-1} \left( \frac{x}{\sqrt{3}} \right) + C$

Yes. And you can check by differentiating that.