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Use the comparison theorem to determine if the improper integral is convergent or divergent. and which function did we use

1) integration from $1$ to infinity of $\frac{sin^2x +5}{x}$ I dont know how to use the ocmparison test for this one.

but the next questoin is I need to show if it converges or diverges.

2) integration from $28$ to infinity of $\frac{1}{(x-1)^\frac{2}{3}}$ i know this diverges but I dont know how to show it

3) Explain why integration from $1$ to $28$ of $\frac{1}{(x-1)^\frac{2}{3}}$ is improper. I don't know hwy this is improper when I know if I evaluate it I get the answer to be $9$

4)integration from $1$ to $\infty$ of $\frac{1}{(x-1)^\frac{2}{3}}$ Right here I want to say that it diverges because $p<1$ but we have $\frac{1}{x-1}$ instead of $\frac{1}{x}$

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$1$. For the first one, note that $\dfrac{\sin^2(x)+5}x > \dfrac{5}x$ for all $x \in [1,\infty)$. Hence, compare the integral with $\displaystyle\int_1^{\infty} \dfrac5x dx$.

$2$. $$\int \dfrac{dx}{(x-1)^{2/3}} = 3 (x-1)^{1/3} + \text{constant}$$ or compare it with $\displaystyle \int \dfrac{dx}{x-1}$.

$3$. Look at what happens to the integrand as you get close to the lower limit of $1$.

$4$. $$\int \dfrac{dx}{(x-1)^{2/3}} = 3 (x-1)^{1/3} + \text{constant}$$ or compare it with $\displaystyle \int \dfrac{dx}{x-1}$.

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  • $\begingroup$ is there any theorem that says if its 1/(x-1)^p and if p<1 it diverges? $\endgroup$ – MathGeek Mar 22 '13 at 21:11
  • $\begingroup$ @MathGeek You could just integrate it out and see by plugging in the limits. $$\int \dfrac{dx}{(x-1)^p} = \dfrac{(x-1)^{1-p}}{1-p} + \text{ constant}$$ $\endgroup$ – user17762 Mar 22 '13 at 21:15
  • $\begingroup$ so for 3 what would we say i know it converges to 0 at 1 but what about 28 ? $\endgroup$ – MathGeek Mar 22 '13 at 21:20
  • $\begingroup$ @MathGeek For $3$, the integrand diverges at $x=1$, whereas the integral does exist. $\endgroup$ – user17762 Mar 22 '13 at 22:05

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