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

In summary, the conversation discussed the integral ##\int \frac{1}{x^2 + 3} \ dx## and the steps to solve it using the substitution method. The final result was ##\frac{\sqrt{3}}{3} \tan^{-1} \left( \frac{x}{\sqrt{3}} \right) + C## and it was suggested to check the answer by differentiating it.

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

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.

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 )##

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##

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.

## What is the formula for integrating 1/(x^2 + 3) dx?

The formula for integrating 1/(x^2 + 3) dx is ∫(1/(x^2 + 3)) dx = (1/√3)tan^-1(x/√3) + C.

## What is the method for integrating 1/(x^2 + 3) dx?

The method for integrating 1/(x^2 + 3) dx is using the substitution u = x/√3, which transforms the integral into ∫(1/(u^2 + 1)) du. This can then be integrated using the inverse tangent function.

## What is the domain of integration for 1/(x^2 + 3) dx?

The domain of integration for 1/(x^2 + 3) dx is all real numbers except for x = ±√3. This is because the denominator, x^2 + 3, cannot be equal to 0.

## What is the graph of 1/(x^2 + 3) dx?

The graph of 1/(x^2 + 3) dx is a hyperbolic curve that approaches the x-axis as x approaches ±∞. It has a vertical asymptote at x = ±√3 and a horizontal asymptote at y = 0.

## What are some real-life applications of integrating 1/(x^2 + 3) dx?

Some real-life applications of integrating 1/(x^2 + 3) dx include calculating the work done by a force, calculating the area under a curve, and solving differential equations in physics and engineering.