Evaluate the integral: I = integral (x² - 1)/(fourth root
Evaluate the integral: $$ I = \int \frac{x^2 - 1}{\sqrt[4]{x + 5}} \, dx. $$
Explanation of Method
When the numerator is quadratic in $x$ but the root in the denominator is a different linear expression, substitute $u$ for the root's argument, express $x^2-1$ in terms of $u$ by expanding $(u-5)^2-1$, and then divide each resulting term by $u^{1/4}$ to obtain a sum of pure power terms.
- Substitute $u$ for the expression inside the root.
- Rewrite the quadratic numerator entirely in terms of $u$.
- Divide each term by $u^{1/4}$ and integrate using the power rule.
- Combine and substitute back, adding the constant of integration.
Step 1 · Use substitution
We begin by using substitution to simplify the integral. Let:
$$ u = x + 5. $$
Then, the derivative of $u$ with respect to $x$ is:
$$ \frac{du}{dx} = 1 \quad \Rightarrow \quad du = dx. $$
Now express $x^2-1$ in terms of $u$. Since $x = u-5$, we have:
$$ x^2-1 = (u-5)^2-1 = u^2-10u+25-1 = u^2-10u+24. $$
So, the integral becomes:
$$ I = \int \frac{u^2-10u+24}{\sqrt[4]{u}}\,du. $$
Step 2 · Simplify the integrand
We can now express the integrand more clearly:
$$ I = \int \frac{u^2-10u+24}{u^{1/4}}\,du. $$
This is the same as:
$$ I = \int \left(u^{2-1/4} - 10u^{1-1/4} + 24u^{-1/4}\right) du. $$
Simplifying the exponents:
$$ I = \int \left(u^{7/4} - 10u^{3/4} + 24u^{-1/4}\right) du. $$
Step 3 · Integrate each term
First term:
$$ \int u^{7/4}\,du = \frac{u^{11/4}}{11/4} = \frac{4}{11}u^{11/4}. $$
Second term:
$$ \int u^{3/4}\,du = \frac{u^{7/4}}{7/4} = \frac{4}{7}u^{7/4}. $$
Multiply by $-10$:
$$ -10\int u^{3/4}\,du = -10\cdot \frac{4}{7}u^{7/4} = -\frac{40}{7}u^{7/4}. $$
Third term:
$$ \int u^{-1/4}\,du = \frac{u^{3/4}}{3/4} = \frac{4}{3}u^{3/4}. $$
Multiply by $24$:
$$ 24\int u^{-1/4}\,du = 24\cdot \frac{4}{3}u^{3/4} = 32u^{3/4}. $$
Step 4 · Combine the results
Now, we combine the results from the three integrals:
$$ I = \frac{4}{11}u^{11/4} - \frac{40}{7}u^{7/4} + 32u^{3/4} + C. $$
Step 5 · Substitute back $u = x + 5$
Finally, substitute $u = x+5$ back into the result:
$$ I = \frac{4}{11}(x+5)^{11/4} - \frac{40}{7}(x+5)^{7/4} + 32(x+5)^{3/4} + C. $$
Summary Table
| Term | Antiderivative |
|---|---|
| $u^{7/4}$ | $\tfrac{4}{11}u^{11/4}$ |
| $-10u^{3/4}$ | $-\tfrac{40}{7}u^{7/4}$ |
| $24u^{-1/4}$ | $32u^{3/4}$ |
Additional Notes
- The domain of $I$ requires $x > -5$, since $x+5 > 0$ is needed for the fourth root to be real and nonzero (the $u^{-1/4}$ term also requires $u\neq 0$).
- Expanding $(u-5)^2-1$ fully before dividing by $u^{1/4}$ avoids having to integrate a product of a polynomial and a root directly.
- The result can be checked by differentiating $\tfrac{4}{11}(x+5)^{11/4}-\tfrac{40}{7}(x+5)^{7/4}+32(x+5)^{3/4}$ and confirming it returns $\dfrac{x^2-1}{\sqrt[4]{x+5}}$.
Result
The solution to the integral is: