The problem asks us to identify which of the given improper integrals converges. A. $\int_{1}^{\infty} \frac{1}{\sqrt[5]{x}} dx$ B. $\int_{0}^{1} x^{-2} dx$ C. $\int_{0}^{\infty} \frac{1}{x+5} dx$ D. $\int_{3}^{5} \frac{1}{\sqrt{x-3}} dx$

1. Problem Description

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

A. $\int_{1}^{\infty} \frac{1}{\sqrt[5]{x}} dx$

B. $\int_{0}^{1} x^{-2} dx$

C. $\int_{0}^{\infty} \frac{1}{x+5} dx$

D. $\int_{3}^{5} \frac{1}{\sqrt{x-3}} dx$

2. Solution Steps

We need to evaluate each integral to determine if it converges or diverges.

A. $\int_{1}^{\infty} \frac{1}{\sqrt[5]{x}} dx = \int_{1}^{\infty} x^{-\frac{1}{5}} dx$

This is an improper integral of the form $\int_{1}^{\infty} x^{-p} dx$, which converges if $p > 1$ and diverges if $p \leq 1$. Here, $p = \frac{1}{5} < 1$, so the integral diverges.

B. $\int_{0}^{1} x^{-2} dx = \lim_{a \to 0^{+}} \int_{a}^{1} x^{-2} dx = \lim_{a \to 0^{+}} [-x^{-1}]_{a}^{1} = \lim_{a \to 0^{+}} [-1 + \frac{1}{a}]$

Since $\lim_{a \to 0^{+}} \frac{1}{a} = \infty$, the integral diverges.

C. $\int_{0}^{\infty} \frac{1}{x+5} dx = \lim_{b \to \infty} \int_{0}^{b} \frac{1}{x+5} dx = \lim_{b \to \infty} [\ln|x+5|]_{0}^{b} = \lim_{b \to \infty} [\ln(b+5) - \ln(5)]$

Since $\lim_{b \to \infty} \ln(b+5) = \infty$, the integral diverges.

D. $\int_{3}^{5} \frac{1}{\sqrt{x-3}} dx = \lim_{a \to 3^{+}} \int_{a}^{5} \frac{1}{\sqrt{x-3}} dx = \lim_{a \to 3^{+}} \int_{a}^{5} (x-3)^{-\frac{1}{2}} dx$

$= \lim_{a \to 3^{+}} [2\sqrt{x-3}]_{a}^{5} = \lim_{a \to 3^{+}} [2\sqrt{5-3} - 2\sqrt{a-3}] = \lim_{a \to 3^{+}} [2\sqrt{2} - 2\sqrt{a-3}] = 2\sqrt{2} - 2\sqrt{0} = 2\sqrt{2}$

The integral converges to $2\sqrt{2}$.

3. Final Answer

D