Evaluate the integral: I = integral dx/(x² + 3)

Evaluate the integral: $$ I = \int \frac{dx}{x^2 + 3}. $$

Explanation of Method

The denominator is already in the standard form $x^2+a^2$, but here $a^2=3$ gives an irrational $a=\sqrt3$. The same formula $\int \frac{dx}{x^2+a^2}=\frac1a\arctan\left(\frac{x}{a}\right)$ still applies, simply carrying the radical through the result.

Recognize the denominator as already matching the standard form $x^2+a^2$.
Identify $a$ from $a^2$, noting it may be irrational.
Apply the standard arctangent integral formula directly.

Step 1 · Recognize the standard form

We recognize that this integral has the standard form:

$$ \int \frac{dx}{x^2+a^2} = \frac{1}{a}\arctan\left(\frac{x}{a}\right). $$

In our case, $a^2 = 3$, so $a = \sqrt{3}$.

Step 2 · Apply the standard result

Using the standard result for the integral, we get:

$$ I = \frac{1}{\sqrt{3}}\arctan\left(\frac{x}{\sqrt{3}}\right) + C, $$

where $C$ is the constant of integration.

Summary Table

Quantity	Value
$a^2$	$3$
$a$	$\sqrt3$
Antiderivative	$\tfrac{1}{\sqrt3}\arctan(x/\sqrt3)$

Additional Notes

The integrand is defined for all real $x$, since $x^2+3 \geq 3 > 0$ always.
Unlike Problem 39, where $a=3$ was an integer, here $a=\sqrt3$ is irrational — the formula applies identically regardless, since it only requires $a>0$.
The result can be checked by differentiating $\tfrac{1}{\sqrt3}\arctan(x/\sqrt3)$ with the chain rule and confirming it returns $\dfrac{1}{x^2+3}$.

Result

The solution to the integral is:

$$ I = \frac{1}{\sqrt{3}}\arctan\left(\frac{x}{\sqrt{3}}\right) + C. $$