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

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

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

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

Homework Helper
Gold Member
2022 Award
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 )##