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Chapter 7: Problem 25

Evaluate the following integrals or state that they diverge. $$\int_{1}^{\infty} \frac{3 x^{2}+1}{x^{3}+x} d x$$

Short Answer

Expert verified
Answer: The integral diverges.

Step by step solution

01

Simplify the integrand

Divide each term in the numerator by \(x^3\), the highest power of \(x\) in the denominator: $$\frac{3 x^{2}+1}{x^{3}+x}=\frac{3\frac{x^2}{x^3}+\frac{1}{x^3}}{\frac{x^3}{x^3}+\frac{x}{x^3}}=\frac{3\frac{1}{x}+\frac{1}{x^3}}{1+\frac{1}{x^2}}$$
02

Integrate using substitution

Let's do a substitution. Let \(u = 1+\frac{1}{x^2}\), then \(-\frac{2}{x^3}dx = du\), so \(x^3 = -\frac{2}{u}\). Now, our integral becomes: $$\int \frac{3\frac{1}{x}+\frac{1}{x^3}}{1+\frac{1}{x^2}}d x = \int \frac{3\frac{1}{x}-\frac{1}{x^3}}{u}\cdot -\frac{2}{x^3}dx = -6\int \frac{(1-\frac{2}{x^2})}{u}d u$$
03

Evaluate the integral

Now, we can rewrite the integral and evaluate it: $$-6\int \frac{(1-\frac{2}{x^2})}{u}d u = -6 \int \left(\frac{1}{u} - \frac{2}{x^2u}\right) du = -6\int \left(\frac{x^2-2}{x^2u}\right) du$$ $$ = -6 \left[ (\ln(u))\Big|_{u = 1}^{\infty} - 2\ln(u)\Big|_{u =1}^{\infty} \right] = -6 \left[ (\ln(\infty) - \ln(1)) - 2(\ln(\infty) - \ln(1)) \right]$$ Since \(\ln(\infty)\) is not defined and the result involves the subtraction of infinity from infinity, we conclude that the integral diverges.

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