Question

$\int \frac{dx}{\sqrt[3]{x+1}+1}$

Answer 1

The given integral diverges.

Answer 2

Solution Explanation

1. Substitution: Let
     $u = \sqrt[3]{x+1}$ so that $x = u^3 - 1$ and $dx = 3u^2\,du$.
     When $x = 1$, $u = \sqrt[3]{2}$; when $x \to \infty$, $u \to \infty$.

2. Change of Variable: The integral becomes
     $\int_{1}^{\infty} \frac{dx}{\sqrt[3]{x+1}+1} = 3\int_{\sqrt[3]{2}}^{\infty}\frac{u^2}{u+1}\,du.$

3. Long Division: Divide $u^2$ by $u+1$
     $\frac{u^2}{u+1} = u - 1 + \frac{1}{u+1}.$

4. Separate and Integrate:
     $3\int_{\sqrt[3]{2}}^\infty \left(u - 1 + \frac{1}{u+1}\right)du.$
     Integrate term-by-term:
     $\int \left(u-1\right)du = \frac{u^2}{2} - u,\quad \int \frac{du}{u+1} = \ln|u+1|.$

5. Check for Convergence:
     Evaluate the antiderivative as $u\to\infty$:
     $\frac{u^2}{2} - u + \ln|u+1| \to \infty.$
     Since the dominating term $\frac{u^2}{2}$ tends to infinity, the integral diverges.

Answer
The given integral diverges.