Evaluate the integral: I = integral (x dx)/(3 - 2x²)
Evaluate the integral: $$ I = \int \frac{x\,dx}{3 - 2x^2}. $$
Explanation of Method
When the numerator is $x$ and the denominator is a quadratic in $x^2$, substituting $u$ for the denominator makes $du$ proportional to $x\,dx$, reducing the integral to the standard logarithmic form $\int \frac{du}{u}$.
- Substitute $u$ for the denominator.
- Match $x\,dx$ to a constant multiple of $du$.
- Integrate using the standard logarithmic antiderivative.
- Substitute back and add the constant of integration.
Step 1 · Set up the substitution
To solve this integral, we use the substitution method. Observing the denominator, we choose:
$$ u = 3 - 2x^2. $$
Now, differentiate both sides with respect to $x$:
$$ \frac{du}{dx} = -4x. $$
Rearranging, we get:
$$ du = -4x\,dx. $$
This allows us to express $x\,dx$ in terms of $du$:
$$ x\,dx = -\frac{du}{4}. $$
Step 2 · Rewrite and integrate
Substituting into the integral:
$$ I = \int \frac{x\,dx}{3-2x^2} = \int \frac{-\frac{du}{4}}{u}. $$
Factor out the constant:
$$ I = -\frac{1}{4}\int \frac{du}{u}. $$
The integral of $\dfrac{1}{u}$ is $\ln|u|$. Thus, we obtain:
$$ I = -\frac{1}{4}\ln|u| + C. $$
Step 3 · Substitute back
Finally, substituting back $u = 3-2x^2$, we get:
$$ I = -\frac{1}{4}\ln\left|3-2x^2\right| + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = 3-2x^2$ |
| $x\,dx$ in terms of $du$ | $-\tfrac14 du$ |
| Antiderivative | $-\tfrac14\ln|u|$ |
Additional Notes
- The domain of $I$ requires $x \neq \pm\sqrt{3/2}$, where the denominator vanishes.
- This pattern — numerator proportional to the derivative of the denominator — is the general signal for a logarithmic antiderivative; here $x\,dx$ is proportional to $du=-4x\,dx$, so the result is $\ln|u|$ scaled by the proportionality constant.
- The result can be checked by differentiating $-\tfrac14\ln\left|3-2x^2\right|$ with the chain rule and confirming it returns $\dfrac{x}{3-2x^2}$.
Result
The solution to the integral is: