Evaluate the integral: I = integral dx/(2 - 3x²)
Evaluate the integral: $$ I = \int \frac{dx}{2 - 3x^2}. $$
Explanation of Method
With a difference of the form $a^2-b^2x^2$ in the denominator, the standard antiderivative is $\int \frac{dx}{a^2-b^2x^2} = \frac{1}{2ab}\ln\left|\frac{a+bx}{a-bx}\right|+C$, derived from the plain $a^2-x^2$ logarithmic form via the substitution $u=bx$.
- Identify $a$ and $b$ by matching the denominator to $a^2-b^2x^2$.
- Apply the standard logarithmic integral formula.
- Simplify the resulting constants and add the constant of integration.
Step 1 · Match the standard form
The denominator $2-3x^2$ matches the form $a^2-b^2x^2$ with:
$$ a^2 = 2,\quad b^2 = 3 \quad \Rightarrow \quad a = \sqrt{2},\quad b = \sqrt{3}. $$
Step 2 · Apply the standard result
We recognize that this integral is of the form:
$$ \int \frac{dx}{a^2 - b^2x^2} = \frac{1}{2ab}\ln\left|\frac{a+bx}{a-bx}\right| + C. $$
Substituting $a=\sqrt2$ and $b=\sqrt3$ into the formula, we have:
$$ I = \frac{1}{2\sqrt{2}\sqrt{3}}\ln\left|\frac{\sqrt{2}+\sqrt{3}\,x}{\sqrt{2}-\sqrt{3}\,x}\right| + C. $$
Step 3 · Simplify the constant
Since $2\sqrt2\sqrt3 = 2\sqrt6$, the integral becomes:
$$ I = \frac{1}{2\sqrt{6}}\ln\left|\frac{\sqrt{2}+\sqrt{3}\,x}{\sqrt{2}-\sqrt{3}\,x}\right| + C. $$
Summary Table
| Quantity | Value |
|---|---|
| $a$, $b$ | $\sqrt2,\ \sqrt3$ |
| Coefficient $\tfrac{1}{2ab}$ | $\tfrac{1}{2\sqrt6}$ |
| Antiderivative | $\tfrac{1}{2\sqrt6}\ln\left|\tfrac{\sqrt2+\sqrt3x}{\sqrt2-\sqrt3x}\right|$ |
Additional Notes
- The domain of $I$ requires $x \neq \pm\sqrt{2/3}$, where the denominator $2-3x^2$ vanishes.
- Deriving the formula $\int\frac{dx}{a^2-b^2x^2}=\frac{1}{2ab}\ln\left|\frac{a+bx}{a-bx}\right|+C$ from the simpler $\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right|+C$ requires substituting $u=bx$, which introduces the extra factor of $\frac{1}{b}$ — dropping that factor is a common source of error in problems like this one.
- The result can be checked by differentiating $\tfrac{1}{2\sqrt6}\ln\left|\tfrac{\sqrt2+\sqrt3x}{\sqrt2-\sqrt3x}\right|$ and confirming it returns $\dfrac{1}{2-3x^2}$.
Result
The solution to the integral is: