Problem 287
Evaluate the integral: I = integral dx/(cube root
Evaluate the integral: $$ I = \int \frac{dx}{\sqrt[3]{3x+1}-1}. $$
Explanation of Method
Substituting $u=\sqrt[3]{3x+1}$ clears the cube root, and polynomial long division resolves the resulting improper rational function.
- Substitute $u=\sqrt[3]{3x+1}$.
- Divide $u^2$ by $(u-1)$.
- Integrate and back-substitute.
Step 1 · Substitute
Let $u=\sqrt[3]{3x+1}$, so $x=\dfrac{u^3-1}3$ and $dx=u^2\,du$:
$$ I = \int\frac{u^2\,du}{u-1}. $$
Step 2 · Divide and integrate
$$ \frac{u^2}{u-1} = u+1+\frac1{u-1} \quad\Rightarrow\quad \int\left(u+1+\frac1{u-1}\right)du = \frac{u^2}2+u+\ln|u-1|+C. $$
Step 3 · Back-substitute
$$ I = \frac{(3x+1)^{2/3}}2 + (3x+1)^{1/3} + \ln\left|\sqrt[3]{3x+1}-1\right| + C. $$
Summary Table
| Piece | Antiderivative |
|---|---|
| $u+1$ | $u^2/2+u$ |
| $1/(u-1)$ | $\ln|u-1|$ |
Additional Notes
- The domain excludes $x=0$, where $\sqrt[3]{3x+1}=1$.
- The result was verified by differentiation: with $w=\sqrt[3]{3x+1}$, the derivative of the boxed expression simplifies to $\dfrac{[(w+1)(w-1)+1]}{w^2(w-1)}=\dfrac{w^2}{w^2(w-1)}=\dfrac1{w-1}$, matching the integrand.
Result
The solution to the integral is:
$$ I = \frac{\sqrt[3]{(3x+1)^2}}2 + \sqrt[3]{3x+1} + \ln\left|\sqrt[3]{3x+1}-1\right| + C. $$