Evaluate the integral: I = integral x² sqrt(x³ - 9) dx
Evaluate the integral: $$ I = \int x^2 \sqrt{x^3 - 9} \, dx. $$
Explanation of Method
Since the derivative of $x^3-9$ is $3x^2$, which is proportional to the $x^2$ factor already present, substituting $u=x^3-9$ converts the square root into $\sqrt{u}$ and reduces the whole problem to a routine power-rule integral.
- Substitute $u$ for the cubic expression under the root.
- Match $x^2\,dx$ to a constant multiple of $du$.
- Integrate the resulting fractional power of $u$.
- 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 expression inside the square root, we choose:
$$ u = x^3 - 9. $$
Now, differentiate both sides with respect to $x$:
$$ \frac{du}{dx} = 3x^2. $$
Rearranging, we get:
$$ du = 3x^2\,dx. $$
This allows us to express $x^2\,dx$ in terms of $du$:
$$ x^2\,dx = \frac{du}{3}. $$
Step 2 · Rewrite and integrate
Substituting into the integral:
$$ I = \int \sqrt{u}\cdot \frac{du}{3}. $$
Factor out the constant:
$$ I = \frac{1}{3}\int u^{1/2}\,du. $$
The integral of $u^{1/2}$ is:
$$ \int u^{1/2}\,du = \frac{2}{3}u^{3/2}. $$
Thus, we obtain:
$$ I = \frac{1}{3}\times \frac{2}{3}u^{3/2} + C = \frac{2}{9}u^{3/2} + C. $$
Step 3 · Substitute back
Finally, substituting back $u = x^3-9$, we get:
$$ I = \frac{2}{9}\left(x^3-9\right)^{3/2} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = x^3-9$ |
| $x^2\,dx$ in terms of $du$ | $\tfrac13 du$ |
| Antiderivative | $\tfrac29 u^{3/2}$ |
Additional Notes
- The domain of $I$ requires $x^3 \geq 9$, i.e. $x \geq \sqrt[3]{9}$, so that $x^3-9\geq0$ under the square root.
- The factor $x^2$ in the integrand is exactly $\tfrac13$ of $\tfrac{d}{dx}(x^3-9)$, which is what makes this substitution work cleanly with no leftover $x$ terms.
- The result can be checked by differentiating $\tfrac29(x^3-9)^{3/2}$ with the chain rule and confirming it returns $x^2\sqrt{x^3-9}$.
Result
The solution to the integral is: