The integral test is a vital calculus method for testing series via improper integrals. For $∑_{n=1}$^∞ $1/(n^2$ + $1)^{3/4}$, f(x) = $1/(x^2$ + $1)^{3/4}$ is positive, continuous, and decreasing. Asymptotically, it resembles ∫ $dx/x^{3/2}$, with 3/2 > 1, so the integral converges. Thus, the series converges. A distractor claims divergence because ∫ $dx/x^{3/2}$ diverges, but it actually converges. Focus on the dominant term for large x to apply the integral test effectively.

The integral test offers a key method in calculus to test series convergence via improper integrals. For $∑_{n=1}$^∞ 1/√n, f(x) = 1/√x is positive, continuous, and decreasing for x ≥ 1. The integral ∫_1^∞ dx/√x = [2√x]_1^∞ diverges to infinity since the exponent 1/2 < 1. Thus, the series diverges. A distractor claims convergence based on ∫ $dx/x^{3/2}$, but that converges while the actual integral diverges. Compare exponents to 1 in p-series equivalents for effective integral test application.

The integral test serves as an essential technique in calculus to evaluate series convergence via improper integrals. For $∑_{n=2}$^∞ 1/(n ln n), use f(x) = 1/(x ln x), positive, continuous, and decreasing for x ≥ 2. The integral ∫2^∞ dx/(x ln x) substitutes u = ln x, du = dx/x, resulting in $∫{ln 2}$^∞ du/u, which is ln u from ln 2 to ∞, diverging to infinity. Therefore, the series diverges because the integral diverges. A distractor might propose convergence based on ∫ dx/(x (ln $x)^2$), but that converges while the given integral does not. To apply the integral test reliably, confirm the function's monotonicity and compute the integral's limit carefully.

The integral test is a key calculus method for determining series behavior via integrals. For $∑_{n=2}$^∞ 1/(n ln n (ln ln $n)^2$), f(x) = 1/(x ln x (ln ln $x)^2$) is positive, continuous, and decreasing for large x. Substituting v = ln ln x gives ∫ $dv/v^2$, which converges to a finite value. Hence, the series converges. A distractor suggests divergence based on ∫ dx/(x ln x), but the additional terms ensure convergence. Use nested substitutions for multi-level logarithms in the integral test.

The integral test is an important tool in calculus for evaluating series through integrals. For $∑_{n=1}$^∞ $1/(n^2$ + 1), f(x) = $1/(x^2$ + 1) is positive, continuous, and decreasing. The integral ∫_1^∞ $dx/(x^2$ + 1) = arctan x to ∞ converges to π/4. Therefore, the series converges. A distractor suggests divergence because ∫ dx/x diverges, but that's irrelevant here. Match the integrand precisely to the series for correct integral test application.

This problem requires using the integral test to determine convergence of the series. We need to evaluate $\int_1^{\infty}\frac{1}{\sqrt{x^3+1}},dx$. For large $x$, the integrand behaves like $\frac{1}{\sqrt{x^3}} = \frac{1}{x^{3/2}}$. Since $\int_1^{\infty}\frac{1}{x^{3/2}},dx = \left[-\frac{2}{\sqrt{x}}\right]_1^{\infty} = 0 - (-2) = 2$ converges, and our integrand is comparable, the original integral also converges. Therefore, by the integral test, the series converges. Choice D incorrectly claims the antiderivative is $\sqrt{x^3+1}$, which would incorrectly suggest divergence. The key strategy is to compare the integrand's behavior to known convergent or divergent integrals for large $x$.

This problem tests the integral test for series convergence. For large $x$, the integrand $\frac{1}{\sqrt{x^2+9}}$ behaves like $\frac{1}{\sqrt{x^2}} = \frac{1}{x}$. We know that $\int_1^{\infty}\frac{1}{x},dx = \ln x|_1^{\infty} = \infty$, which diverges. By comparison, since $\frac{1}{\sqrt{x^2+9}} \sim \frac{1}{x}$ for large $x$, the integral $\int_1^{\infty}\frac{1}{\sqrt{x^2+9}},dx$ also diverges. Choice A incorrectly claims convergence, possibly confusing this with cases where the denominator grows faster than linearly. For integrands of the form $\frac{1}{\sqrt{x^2+c}}$, the integral behaves like $\int\frac{1}{x},dx$ and thus diverges.

This problem requires using the Integral Test to determine whether the series converges. The Integral Test states that if f(x) is positive, continuous, and decreasing for x ≥ 2, then the series and the improper integral have the same convergence behavior. For f(x) = 1/(x(ln x)²), we evaluate ∫₂^∞ 1/(x(ln x)²) dx using the substitution u = ln x, du = dx/x, which gives $∫{ln 2}$^∞ 1/u² du = $[-1/u]{ln 2}$^∞ = 0 - (-1/ln 2) = 1/ln 2, a finite value. Since the integral converges, the series also converges. Choice C incorrectly evaluates the integral as 1/(ln x)² without proper integration, while choice D incorrectly computes the antiderivative. When applying the Integral Test, always verify that f(x) satisfies all conditions and carefully evaluate the improper integral using appropriate techniques.

This problem applies the Integral Test to ∑ 1/(n√n) = ∑ n^(-3/2). The function f(x) = x^(-3/2) is positive, continuous, and decreasing for x ≥ 1. To evaluate ∫₁^∞ x^(-3/2) dx, we use the power rule: ∫ x^(-3/2) dx = x^(-1/2)/(-1/2) = -2x^(-1/2) + C = -2/√x + C. Computing the improper integral: [-2/√x]₁^∞ = 0 - (-2) = 2, a finite value. Since the integral converges, the series also converges. Choice D has the correct antiderivative but incorrectly claims divergence, while choice C suggests an incorrect antiderivative of ln x. For p-series ∑ $1/n^p$, convergence occurs when p > 1, and here p = 3/2 > 1, confirming convergence.

This problem applies the integral test to a quadratic denominator series. We evaluate $\int_1^{\infty}\frac{1}{(2x+1)^2},dx$ using substitution $u = 2x+1$, so $du = 2dx$. The integral becomes $\frac{1}{2}\int_3^{\infty}\frac{1}{u^2},du = \frac{1}{2}\left[-\frac{1}{u}\right]_3^{\infty} = \frac{1}{2}(0 - (-\frac{1}{3})) = \frac{1}{6}$. Since the improper integral converges to a finite value, the series converges by the integral test. Choice D shows an incorrect antiderivative missing the factor of $-\frac{1}{2}$ from the chain rule. For integrals involving $(ax+b)^n$ with $n > 1$, always remember to include the factor $\frac{1}{a}$ when substituting.