Evaluate the integral: integral (x² - 3) 5th root of (x³
Evaluate the integral: $$ \int \left(x^2 - 3\right) \sqrt[5]{x^3 - 9x}\,dx. $$
Explanation of Method
The derivative of $x^3-9x$ is $3x^2-9=3(x^2-3)$, which is proportional to the polynomial factor multiplying the root. Substituting $u=x^3-9x$ therefore converts the fifth root into $u^{1/5}$ and reduces the whole problem to a routine power-rule integral.
- Substitute $u$ for the cubic expression under the root.
- Match $(x^2-3)\,dx$ to a constant multiple of $du$.
- Integrate the resulting fractional power of $u$.
- Substitute back and add the constant of integration.
Step 1 · Substitution
We begin by recognizing that the expression inside the fifth root, $x^3-9x$, can be simplified with a substitution. Let:
$$ u = x^3 - 9x. $$
Now, differentiate both sides with respect to $x$:
$$ du = (3x^2-9)\,dx. $$
This allows us to express $(x^2-3)\,dx$ in terms of $du$:
$$ (x^2-3)\,dx = \frac{du}{3}. $$
Step 2 · Substitute into the integral
Substitute $u=x^3-9x$ and $(x^2-3)\,dx=\dfrac{du}{3}$ into the integral:
$$ \int \left(x^2-3\right)\sqrt[5]{x^3-9x}\,dx = \int \frac{1}{3}\sqrt[5]{u}\,du. $$
Step 3 · Simplifying the integral
Now, we can simplify the integral further:
$$ = \frac{1}{3}\int u^{1/5}\,du. $$
Step 4 · Integrate
The integral of $u^{1/5}$ is straightforward:
$$ \int u^{1/5}\,du = \frac{u^{6/5}}{6/5} = \frac{5u^{6/5}}{6}. $$
Thus, the integral becomes:
$$ \frac{1}{3}\cdot \frac{5u^{6/5}}{6} = \frac{5u^{6/5}}{18}. $$
Step 5 · Substitute back $u = x^3 - 9x$
Finally, substitute $u = x^3-9x$ back into the solution:
$$ \frac{5}{18}\left(x^3-9x\right)^{6/5} + C. $$
Summary Table
| Quantity | Value |
|---|---|
| Substitution | $u = x^3-9x$ |
| $(x^2-3)\,dx$ in terms of $du$ | $\tfrac13 du$ |
| Antiderivative | $\tfrac{5}{18}u^{6/5}$ |
Additional Notes
- The integrand is defined for all real $x$, since fifth roots (odd roots) are defined for both positive and negative arguments.
- The polynomial factor $x^2-3$ is exactly $\tfrac13$ of $\tfrac{d}{dx}(x^3-9x)$, which is what makes this substitution collapse cleanly with no leftover $x$ terms.
- The result can be checked by differentiating $\tfrac{5}{18}(x^3-9x)^{6/5}$ with the chain rule and confirming it returns $(x^2-3)\sqrt[5]{x^3-9x}$.
Result
The solution to the integral is: