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The integral is this. Using the comparison method, does this integral converge or diverge?$$\int_{1}^{2}\frac{\sqrt{\left(x^{4}+1\right)}}{\left(x^{3}-1\right)}$$

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3 Answers

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Observe that $$\int_{1}^{2}\frac{\sqrt{\left(x^{4}+1\right)}}{\left(x^{3}-1\right)} > \int_{1}^{2}\frac{1}{\left(x^{3}-1\right)} > \int_{1}^{2}\frac{1}{\left(x^{4}-1\right)}$$

Both the integrals $$\int_{1}^{2}\frac{1}{\left(x^{3}-1\right)} \text{ and } \int_{1}^{2}\frac{1}{\left(x^{4}-1\right)} \text{ do not converge}$$

You can show the divergence of either one (whichever looks easier to you), and complete the proof using comparison test (as asked in the question).

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The function $$f(x)=\frac{\sqrt{x^4+1}}{x^3-1}$$

is continuous on $(1,2]$, so you just need to see what happens near $1$. And near $1$, you have$$\frac{\sqrt{x^4+1}}{x^3-1}=\frac{\sqrt{x^4+1}}{(x-1)(x^2+x+1)} \sim \frac{\sqrt{2}}{3(x-1)}$$

which is not integrable.

Therefore the integral diverges.

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$$I=\int_{a}^{b} \frac{dx}{(x-a)^\alpha}$$is finite (convergent) if $\alpha <1$, divergent otherwise. Therefore,$$\int_{1}^{2}\frac{\sqrt{x^4+1}}{1+x+x^2} ~ \frac{dx}{x-1}$$is divergent as $\alpha=1$.

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