Evaluate the integral: I = integral dx/(2 + cube root
Evaluate the integral: $$ I = \int \frac{dx}{2 + \sqrt[3]{x - 1}}. $$
Explanation of Method
When a root appears added to a constant in a denominator, a single substitution isn't enough to clear the radical — a second substitution setting $v$ equal to the root itself, so that $u = v^3$, turns the denominator into the polynomial $2+v$. Polynomial long division then separates the integrand into power terms plus a logarithmic term.
- Substitute $u$ for the linear expression under the root.
- Substitute again, $v = u^{1/3}$, to clear the radical entirely.
- Divide the resulting rational function of $v$ using polynomial long division.
- Integrate each term, then substitute back through $v$ and $u$ to $x$.
Step 1 · Use substitution
We begin by using substitution to simplify the integral. Let:
$$ u = x - 1. $$
Then, the derivative of $u$ with respect to $x$ is:
$$ \frac{du}{dx} = 1 \quad \Rightarrow \quad du = dx. $$
Now, the integral becomes:
$$ I = \int \frac{du}{2 + \sqrt[3]{u}}. $$
Step 2 · Simplify the integrand
We now need to simplify the integrand. Rewrite $\sqrt[3]{u}$ as $u^{1/3}$, so the integral becomes:
$$ I = \int \frac{du}{2 + u^{1/3}}. $$
To further simplify, let's try the substitution $v = u^{1/3}$, so that $u = v^3$. The derivative of $u$ with respect to $v$ is:
$$ \frac{du}{dv} = 3v^2 \quad \Rightarrow \quad du = 3v^2\,dv. $$
Substituting these into the integral:
$$ I = \int \frac{3v^2\,dv}{2+v}. $$
Step 3 · Use polynomial division
Next, we perform polynomial division on the integrand:
$$ \frac{3v^2}{2+v} = 3v - 6 + \frac{12}{2+v}. $$
Thus, the integral becomes:
$$ I = \int \left(3v - 6 + \frac{12}{2+v}\right) dv. $$
Step 4 · Integrate each term
Now we can integrate each term separately: $\displaystyle \int 3v\,dv = \dfrac{3v^2}{2}$, $\displaystyle \int -6\,dv = -6v$, and $\displaystyle \int \dfrac{12}{2+v}\,dv = 12\ln|2+v|$.
Thus, the integral becomes:
$$ I = \frac{3v^2}{2} - 6v + 12\ln|2+v| + C. $$
Step 5 · Substitute back $v = u^{1/3}$ and $u = x - 1$
Finally, substitute $v = u^{1/3}$ and $u = x-1$ back into the result:
$$ I = \frac{3(u^{1/3})^2}{2} - 6(u^{1/3}) + 12\ln\left|2+u^{1/3}\right| + C. $$
This simplifies to:
$$ I = \frac{3(x-1)^{2/3}}{2} - 6(x-1)^{1/3} + 12\ln\left|2+(x-1)^{1/3}\right| + C. $$
Summary Table
| Term in $v$ | Antiderivative |
|---|---|
| $3v$ | $\tfrac{3}{2}v^2$ |
| $-6$ | $-6v$ |
| $12/(2+v)$ | $12\ln|2+v|$ |
Additional Notes
- The domain of $I$ requires $2+\sqrt[3]{x-1} \neq 0$, i.e. $x \neq 1-8=-7$, since the original denominator must not vanish.
- The double substitution $u=x-1$ then $v=u^{1/3}$ (so $u=v^3$) is the standard technique for clearing a root that appears added to a constant, since it converts the whole denominator into a linear polynomial in $v$.
- The result can be checked by differentiating $\tfrac32(x-1)^{2/3}-6(x-1)^{1/3}+12\ln\left|2+(x-1)^{1/3}\right|$ and confirming it returns $\dfrac{1}{2+\sqrt[3]{x-1}}$.
Result
The solution to the integral is: