The problem asks us to identify which of the given improper integrals converges. A. $\int_{1}^{\infty} \frac{1}{\sqrt[3]{x}} dx$ B. $\int_{0}^{1} x^{-5} dx$ C. $\int_{0}^{\infty} \frac{1}{x+3} dx$ D. $\int_{1}^{4} \frac{1}{\sqrt{x-1}} dx$

1. Problem Description

The problem asks us to identify which of the given improper integrals converges.

A. $\int_{1}^{\infty} \frac{1}{\sqrt[3]{x}} dx$

B. $\int_{0}^{1} x^{-5} dx$

C. $\int_{0}^{\infty} \frac{1}{x+3} dx$

D. $\int_{1}^{4} \frac{1}{\sqrt{x-1}} dx$

2. Solution Steps

A. $\int_{1}^{\infty} \frac{1}{\sqrt[3]{x}} dx = \int_{1}^{\infty} x^{-1/3} dx$.

Since the exponent $-1/3$ is greater than $-1$, this integral diverges. More specifically,

$\int_{1}^{\infty} x^{-1/3} dx = \lim_{b \to \infty} \int_{1}^{b} x^{-1/3} dx = \lim_{b \to \infty} [\frac{3}{2} x^{2/3}]_{1}^{b} = \lim_{b \to \infty} (\frac{3}{2} b^{2/3} - \frac{3}{2}) = \infty$.

B. $\int_{0}^{1} x^{-5} dx$. Since the upper limit is finite, and the singularity occurs at $x=0$, we have $\int_{0}^{1} x^{-5} dx = \lim_{a \to 0^+} \int_{a}^{1} x^{-5} dx = \lim_{a \to 0^+} [\frac{x^{-4}}{-4}]_{a}^{1} = \lim_{a \to 0^+} (-\frac{1}{4} + \frac{1}{4a^4}) = \infty$.

The integral diverges.

C. $\int_{0}^{\infty} \frac{1}{x+3} dx$.

$\int_{0}^{\infty} \frac{1}{x+3} dx = \lim_{b \to \infty} \int_{0}^{b} \frac{1}{x+3} dx = \lim_{b \to \infty} [\ln|x+3|]_{0}^{b} = \lim_{b \to \infty} (\ln|b+3| - \ln|3|) = \infty$.

The integral diverges.

D. $\int_{1}^{4} \frac{1}{\sqrt{x-1}} dx$. The singularity is at $x=1$.

$\int_{1}^{4} \frac{1}{\sqrt{x-1}} dx = \lim_{a \to 1^+} \int_{a}^{4} \frac{1}{\sqrt{x-1}} dx = \lim_{a \to 1^+} [2\sqrt{x-1}]_{a}^{4} = \lim_{a \to 1^+} (2\sqrt{4-1} - 2\sqrt{a-1}) = 2\sqrt{3} - 2\sqrt{0} = 2\sqrt{3}$.

This integral converges to $2\sqrt{3}$.

3. Final Answer

The integral that will converge is D. $\int_{1}^{4} \frac{1}{\sqrt{x-1}} dx$.