Calculate the generalized integral

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Attachment 14470

Am I correct? and then?

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- Dec 15th 2009, 11:13 AMsf1903Calculate the generalized integral
Calculate the generalized integral

Attachment 14469

Attachment 14470

Am I correct? and then? - Dec 15th 2009, 11:33 AMtonio

Sorry to say but you're not even close to correct, and if you wrote the above then I think you need to go back to your general basics on integration:

$\displaystyle \int\frac{x^2-1}{x^4}dx=\int\left(\frac{1}{x^2}-\frac{1}{x^4}\right)dx=-\frac{1}{x}+\frac{1}{3x^3}$.

Now evaluate the above in the integral's limits and make the upper limit tend to $\displaystyle \infty$ ...the answer must be $\displaystyle \frac{2}{3}$

Tonio - Dec 15th 2009, 12:48 PMKrizalid
we can also turn this improper integral into a proper one, so let's put $\displaystyle x=\frac1t$ and the integral becomes $\displaystyle \int_{0}^{1}{\left( 1-t^{2} \right)\,dt}=1-\frac{1}{3}=\frac{2}{3}.$