Evaluate the integral: I = integral dx/(2 + 3x²)
Evaluate the integral: $$ I = \int \frac{dx}{2 + 3x^2}. $$
Explanation of Method
When $x^2$ has a coefficient other than $1$, factor that coefficient out of the denominator first. This leaves a pure $x^2+a^2$ form inside the parentheses, to which the standard arctangent formula applies directly, with the factored constant carried along outside the integral.
- Factor the coefficient of $x^2$ out of the denominator.
- Identify $a$ from the resulting standard form.
- Apply the standard arctangent formula and simplify the constants.
Step 1 · Factor the denominator
We first factor the constant term from the denominator:
$$ I = \int \frac{dx}{3\left(\frac{2}{3}+x^2\right)} = \frac{1}{3}\int \frac{dx}{\frac{2}{3}+x^2}. $$
Step 2 · Apply the standard result
We recognize that this is a standard integral of the form:
$$ \int \frac{dx}{a^2+x^2} = \frac{1}{a}\arctan\left(\frac{x}{a}\right), $$
where $a^2 = \dfrac{2}{3}$, so $a = \sqrt{\dfrac{2}{3}} = \dfrac{\sqrt{6}}{3}$. Substituting this into the integral, we have:
$$ I = \frac{1}{3}\cdot \frac{1}{\frac{\sqrt{6}}{3}}\arctan\left(\frac{x}{\frac{\sqrt{6}}{3}}\right). $$
Simplifying the constant:
$$ I = \frac{1}{\sqrt{6}}\arctan\left(\frac{3x}{\sqrt{6}}\right) + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Factored form | $3\left(\tfrac23+x^2\right)$ |
| $a$ | $\sqrt6/3$ |
| Antiderivative | $\tfrac{1}{\sqrt6}\arctan(3x/\sqrt6)$ |
Additional Notes
- The integrand is defined for all real $x$, since $2+3x^2 \geq 2 > 0$ always.
- The factor $\dfrac{1}{3}$ pulled out front combines with $\dfrac{1}{a}=\dfrac{3}{\sqrt6}$ from the formula to give the overall coefficient $\dfrac{1}{\sqrt6}$; keeping track of both factors separately avoids arithmetic slips.
- The result can be checked by differentiating $\dfrac{1}{\sqrt6}\arctan(3x/\sqrt6)$ with the chain rule and confirming it returns $\dfrac{1}{2+3x^2}$.
Result
The solution to the integral is: