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I'm stuck on finding a good comparison test for this improper integral. Can anyone please help me out?

$$\int_{6}^{7} \frac{(x-4)(3x+1)}{\sqrt{x-6}}dx$$

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  • $\begingroup$ The substitution $x=6+u^2$ makes the problem pretty trivial. The integral of a polynomial over $[0,1]$ is clearly finite. $\endgroup$ – Jack D'Aurizio Mar 12 '18 at 20:31
  • $\begingroup$ @dg123 Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – user Mar 17 '18 at 22:50
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I don't think you need a comparison test. Substitute $u = x-6$ and then divide out the numerator and the denominator to separate and simplify the fraction. 3 of 4 integrals will be bounded obviously, and the last one will be of the form $$\int_0^1 \frac{dx}{\sqrt{x}}$$ which is integrable...

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$x\le7\implies x-4\le3$ and $3x+1\le22$

thus $$\frac {(x-4)(3x+1)}{\sqrt {x-6}}\le\frac {66}{\sqrt {x-6}} $$

the right side integral converges, so...

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HINT

Note that

$$\int_{6}^{7} \frac{(x-4)(3x+1)}{\sqrt{x-6}}dx=\int_{0}^{1} \frac{(y+2)(3y+19)}{\sqrt{y}}dy=\int_{0}^{1} \frac{3y^2+25y+38}{\sqrt{y}}dy$$

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