Evaluate the integral: integral (x² dx)/((8x³ + 27)^(2/3))
Evaluate the integral: $$ \int \frac{x^2\,dx}{\left(8x^3 + 27\right)^{2/3}}. $$
Explanation of Method
The derivative of $8x^3+27$ is $24x^2$, proportional to the numerator $x^2$, so substituting $u=8x^3+27$ reduces the integral to a routine negative-fractional-power antiderivative in $u$.
- Substitute $u$ for the cubic expression.
- Match $x^2\,dx$ to a constant multiple of $du$.
- Integrate using the power rule.
- Substitute back and add the constant of integration.
Step 1 · Substitution
We begin by using substitution to simplify the integral. Let:
$$ u = 8x^3 + 27. $$
Then, differentiate both sides with respect to $x$:
$$ \frac{du}{dx} = 24x^2. $$
So, $du = 24x^2\,dx$. Now, solve for $x^2\,dx$:
$$ x^2\,dx = \frac{du}{24}. $$
Step 2 · Substitute and integrate
Substitute this into the original integral:
$$ \int \frac{x^2\,dx}{(8x^3+27)^{2/3}} = \int \frac{\frac{du}{24}}{u^{2/3}} = \frac{1}{24}\int u^{-2/3}\,du. $$
Now, integrate $u^{-2/3}$:
$$ \int u^{-2/3}\,du = \frac{u^{1/3}}{1/3} = 3u^{1/3}. $$
Thus, the integral becomes:
$$ \frac{1}{24}\cdot 3u^{1/3} = \frac{3}{24}u^{1/3} = \frac{1}{8}u^{1/3}. $$
Step 3 · Substitute back
Finally, substitute $u = 8x^3+27$ back into the solution:
$$ \frac{1}{8}\left(8x^3+27\right)^{1/3} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = 8x^3+27$ |
| $x^2\,dx$ in terms of $du$ | $\tfrac{1}{24}du$ |
| Antiderivative | $\tfrac18 u^{1/3}$ |
Additional Notes
- The domain of $I$ requires $8x^3+27 \neq 0$, i.e. $x \neq -\tfrac32$, where the denominator vanishes.
- Since $24x^2$ is exactly $24$ times the numerator $x^2$, this substitution reduces cleanly with a single constant factor, similar in spirit to Problem 64.
- The result can be checked by differentiating $\tfrac18(8x^3+27)^{1/3}$ with the chain rule and confirming it returns $\dfrac{x^2}{(8x^3+27)^{2/3}}$.
Result
The solution to the integral is: